Proving properties of a parallelogram
A parallelogram is defined by its sides being parallel, and nothing in that says how long they are or where its diagonals cross. What makes opposite sides equal in length, and why do the diagonals cut each other exactly in half?
Where you are 4 in Proofs about parallelograms
When you reach for it
A proof has a parallelogram in it and needs one of two facts the definition does not give. The definition says only that the opposite sides lie on parallel lines. Two segments on two parallel lines can be any lengths at all, and two diagonals drawn across the shape could, for all the definition says, cross anywhere inside it. Both facts have to be earned.
- Opposite sides are congruent. Reach for this when a proof needs one side equal in length to the side opposite, or when a problem gives sides in terms of a letter.
- The diagonals bisect each other. Reach for this when a proof needs the crossing of the diagonals to be a midpoint: to find a diagonal from part of it, to find a missing corner in coordinates, or to show that a segment through the crossing is cut in half there.
These are the first two of the properties of a parallelogram to be proved, and both proofs follow the pattern of proofs about parallelograms: draw a diagonal, let it cut the figure into triangles, show two of the triangles congruent, and read the wanted parts off them with corresponding parts. The first theorem is proved with one diagonal. The second is proved with both, and uses the first as one of its steps. Once proved, the words Opposite Sides Theorem, and “diagonals of a parallelogram bisect each other”, may stand as the reason for one line in any later proof, which is how they are used in the examples below.
The procedure
Proving opposite sides congruent
The theorem. The opposite sides of a parallelogram are congruent. Set out as in writing the Given and the Prove:
Given. ABCD is a parallelogram: \overline{AB} \parallel \overline{DC} and \overline{AD} \parallel \overline{BC}.
Prove. \overline{AB} \cong \overline{CD} and \overline{BC} \cong \overline{DA}.
The letters go round the figure, so \overline{AB} and \overline{CD} are opposite, and \overline{AD} and \overline{BC} are opposite. \overline{AC} joins two opposite corners and is not a side; it is a diagonal.
Plan. The Given is about angles in disguise, because parallel lines make congruent angles with any line that crosses them. The Prove is about lengths. The bridge from angles to lengths is a pair of congruent triangles, and a triangle needs a third side. So draw a diagonal, which is an auxiliary line that the problem did not hand over. The diagonal \overline{AC} crosses both pairs of parallel sides, so one line supplies two pairs of angles, and it is itself a side of both triangles.
Proof.
| Statements | Reasons |
|---|---|
| 1. ABCD is a parallelogram | given |
| 2. \overline{AB} \parallel \overline{DC} | definition of parallelogram: line 1 |
| 3. \overline{AD} \parallel \overline{BC} | definition of parallelogram: line 1 |
| 4. Draw the diagonal \overline{AC} | two points determine a line |
| 5. \angle BAC \cong \angle DCA | Alternate Interior Angles Theorem: line 2, transversal \overline{AC} |
| 6. \angle BCA \cong \angle DAC | Alternate Interior Angles Theorem: line 3, transversal \overline{AC} |
| 7. \overline{AC} \cong \overline{CA} | reflexive property of congruence |
| 8. \triangle ABC \cong \triangle CDA | ASA: lines 5, 7 and 6 |
| 9. \overline{AB} \cong \overline{CD} | corresponding parts of congruent triangles: line 8 |
| 10. \overline{BC} \cong \overline{DA} | corresponding parts of congruent triangles: line 8 |
Each line earns its place as follows.
Lines 5 and 6. The Alternate Interior Angles Theorem says that when parallel lines are crossed by a transversal, two angles that lie between the lines, one at each crossing and on opposite sides of the transversal, are congruent (alternate interior angles). For line 5 the parallel lines are \overleftrightarrow{AB} and \overleftrightarrow{DC}. The transversal is \overleftrightarrow{AC}. The angles \angle BAC and \angle DCA lie between those lines, with B and D on opposite sides of the diagonal, a position fact the figure shows (what a diagram may assume). Line 6 is the same move for the other pair of parallel sides, which is why the Given has two parts and the proof uses both.
Line 7. The diagonal is a side of triangle ABC and also a side of triangle CDA, the very same segment. A segment is congruent to itself.
Line 8, and the order of the letters. ASA needs two angles and the side between them. In \triangle ABC the side \overline{AC} joins A and C, so it lies between \angle BAC and \angle BCA. In \triangle CDA the side \overline{CA} lies between \angle DCA and \angle DAC. Line 5 pairs the angle at A with the angle at C, and line 6 pairs the angle at C with the angle at A, so the vertices correspond as A \to C, B \to D, C \to A. The congruence statement is written in exactly that order, \triangle ABC \cong \triangle CDA, because the order of the letters is what says which parts match (congruence statements and correspondence). Writing \triangle ADC for the second triangle would pair A with A and say something false.
Lines 9 and 10. With A \to C, B \to D, C \to A, the side \overline{AB} corresponds to \overline{CD}, and \overline{BC} to \overline{DA}. Those are the two pairs of opposite sides.
The same proof in a paragraph. Draw the diagonal \overline{AC}. Because \overline{AB} \parallel \overline{DC} and \overline{AC} crosses them, \angle BAC \cong \angle DCA, and because \overline{AD} \parallel \overline{BC}, \angle BCA \cong \angle DAC. The two triangles share \overline{AC}, so by ASA they are congruent, and their corresponding sides give \overline{AB} \cong \overline{CD} and \overline{BC} \cong \overline{DA}.
Either diagonal. Drawing \overline{BD} instead works the same way, with \triangle ABD and \triangle CDB. Nothing in the argument favored \overline{AC}.
From congruent to equal. The theorem says the sides are congruent. Congruent segments have equal lengths, so the next step in a numerical problem is to write AB = CD and use it.
Proving the diagonals bisect each other
Draw a parallelogram, draw both diagonals, and measure. The point where they cross always sits halfway along each one. In Figure 3 that means AE = CE and BE = DE.
The theorem. The diagonals of a parallelogram bisect each other.
“Bisect each other” is two claims at once. \overline{BD} cuts \overline{AC} into two congruent halves, so \overline{AE} \cong \overline{CE}, and \overline{AC} cuts \overline{BD} into two congruent halves, so \overline{BE} \cong \overline{DE}. Both say that E is the midpoint of a diagonal. The theorem is also the reason a parallelogram looks the same after a half turn about E: the turn swaps A with C and B with D, because E is the middle of both.
Given. ABCD is a parallelogram, so both pairs of opposite sides are parallel, and its diagonals meet at E. Everything else has to be earned.
Prove. \overline{AE} \cong \overline{CE} and \overline{BE} \cong \overline{DE}.
Choose the triangles that hold the halves. The Prove is about the four segments \overline{AE}, \overline{CE}, \overline{BE} and \overline{DE}. For a pair of congruent triangles to hand them over, \overline{AE} and \overline{CE} must be matching sides of two triangles. The two triangles opposite each other across E, \triangle ABE at the bottom and \triangle CDE at the top, do it: \overline{AE} is a side of one, \overline{CE} of the other, and the same pair also holds \overline{BE} and \overline{DE}. The triangles made by one diagonal, \triangle ABC and \triangle CDA, will not do here: they share the whole of \overline{AC} and neither has half of a diagonal as a side.
Get two angles from the parallel sides. Each diagonal is a transversal of the parallel sides \overline{AB} and \overline{DC}. Along \overline{AC}, the angles \angle BAE and \angle DCE lie between the parallels and on opposite sides of the transversal, so they are alternate interior angles and are congruent. Along \overline{BD}, the same holds for \angle ABE and \angle CDE. Figure 4 marks both pairs.
Get the side from the Opposite Sides Theorem. The side between those angles is \overline{AB} in one triangle and \overline{CD} in the other, and they are congruent by the theorem just proved. That theorem was proved with one diagonal, through \triangle ABC \cong \triangle CDA, and says nothing about E, so citing it here does not lean on what is being proved. If a course has not yet proved it, prove it inside the argument, as Example 10 does.
Prove the triangles congruent. Two angles and the side between them is ASA, one of the congruence criteria. Check that the side really is between the angles: in \triangle ABE the angles are at A and B, and \overline{AB} joins them. The statement \triangle ABE \cong \triangle CDE pairs A \leftrightarrow C, B \leftrightarrow D and E \leftrightarrow E, which is the pairing the marks show.
Claim both halves, then name the midpoint. Under that pairing \overline{AE} corresponds to \overline{CE} and \overline{BE} to \overline{DE}, so one congruence gives both claims. E lies on both diagonals, so it is the midpoint of each, and the definition of a segment bisector finishes the proof.
| Statements | Reasons |
|---|---|
| 1. ABCD is a parallelogram with diagonals \overline{AC} and \overline{BD} meeting at E | given |
| 2. \overline{AB} \parallel \overline{DC} | definition of parallelogram: line 1 |
| 3. \angle BAE \cong \angle DCE | Alternate Interior Angles Theorem: line 2, transversal \overline{AC} |
| 4. \angle ABE \cong \angle CDE | Alternate Interior Angles Theorem: line 2, transversal \overline{BD} |
| 5. \overline{AB} \cong \overline{CD} | Opposite Sides Theorem: line 1 |
| 6. \triangle ABE \cong \triangle CDE | ASA postulate: lines 3, 5 and 4 |
| 7. \overline{AE} \cong \overline{CE} and \overline{BE} \cong \overline{DE} | corresponding parts of congruent triangles: line 6 |
| 8. E is the midpoint of \overline{AC} and of \overline{BD} | definition of midpoint: line 7, E is on both diagonals (line 1) |
| 9. \overline{AC} and \overline{BD} bisect each other | definition of segment bisector: line 8 |
Line 2 uses only one pair of parallel sides. The other pair is not wasted: it is hidden inside line 5, since the proof that opposite sides are congruent needs both pairs.
Worked examples
Sides written in letters
In Figure 5, ABCD is a parallelogram with AB = 3x + 4, CD = 5x - 8, BC = 2y + 1 and AD = y + 9. Find x, y and the perimeter.
\overline{AB} and \overline{CD} are opposite sides of a parallelogram, so by the Opposite Sides Theorem they are congruent, and their lengths are equal:
So AB = 3(6) + 4 = 22. The other pair, \overline{BC} and \overline{AD}, is opposite too:
So BC = 2(8) + 1 = 17. The perimeter is 22 + 17 + 22 + 17 = 78.
Check. CD = 5(6) - 8 = 22 and AD = 8 + 9 = 17, matching their opposite sides.
A perimeter with one unknown
A parallelogram has perimeter 58 cm, and one side is 5 cm longer than a side next to it. Find all four sides.
SolutionCall the shorter side s cm, so its neighbor is s + 5 cm. Without the theorem the four sides would be four unknowns. With it, the side opposite s is s and the side opposite s + 5 is s + 5, so the perimeter is
Set this equal to 58: 4s = 48, so s = 12. The sides are 12, 17, 12 and 17 cm.
Check. 12 + 17 + 12 + 17 = 58.
Lengths from the halves
ABCD is a parallelogram whose diagonals meet at E. AE = 3x + 2, CE = 5x - 10 and BD = 26. Find x, AC and BE.
SolutionThe diagonals bisect each other, so E is the midpoint of \overline{AC} and the two halves have equal lengths:
Then AE = 3(6) + 2 = 20, and AC is two halves, so AC = 40.
E is also the midpoint of \overline{BD}, so BE is half of BD: BE = 26 \div 2 = 13.
Check. CE = 5(6) - 10 = 20, equal to AE as it must be. Nothing ties AE to BE: the diagonals are cut in half, but they need not be the same length.
Two unknowns across both diagonals
ABCD is a parallelogram whose diagonals meet at E. AE = x + y, CE = 14, BE = x - y and DE = 4. Find x, y, AC and BD.
SolutionEach diagonal is bisected, and each bisection gives one equation. From \overline{AC}, AE = CE; from \overline{BD}, BE = DE:
Adding the two equations removes y: 2x = 18, so x = 9. Then 9 + y = 14 gives y = 5.
So AC = 2(14) = 28 and BD = 2(4) = 8.
Check. AE = 9 + 5 = 14 and BE = 9 - 5 = 4, matching CE and DE. Setting AE equal to BE would have been wrong: those halves belong to different diagonals.
The fourth corner in coordinates
A(1, 1), B(7, 2) and C(9, 6) are three vertices of parallelogram ABCD, named in order around it. Find D.
SolutionThe diagonals are \overline{AC} and \overline{BD}, and they bisect each other, so they have the same midpoint. The midpoint formula gives the midpoint of \overline{AC}:
E must also be the midpoint of \overline{BD}. Write D = (x, y) and match each coordinate:
so 7 + x = 10 and 2 + y = 7, giving x = 3 and y = 5. The fourth vertex is D(3, 5).
Check. From A to B is 6 right and 1 up, and from D(3, 5) to C(9, 6) is also 6 right and 1 up. From A to D is 2 right and 4 up, and from B to C is also 2 right and 4 up. Both pairs of opposite sides have matching slopes, so ABCD really is a parallelogram.
New letters and a tilted figure
In Figure 6, PQRS is a parallelogram with \overline{PQ} \parallel \overline{SR} and \overline{PS} \parallel \overline{QR}. Prove that \overline{PQ} \cong \overline{RS} and \overline{PS} \cong \overline{RQ}.
Given. PQRS is a parallelogram.
Prove. \overline{PQ} \cong \overline{RS} and \overline{PS} \cong \overline{RQ}.
Plan. The same as the proof of the Opposite Sides Theorem, with the diagonal \overline{QS} this time. The letters and the tilt are new, but the proof has the same shape. Watch only the order of the letters.
| Statements | Reasons |
|---|---|
| 1. PQRS is a parallelogram | given |
| 2. \overline{PQ} \parallel \overline{SR} | definition of parallelogram: line 1 |
| 3. \overline{PS} \parallel \overline{QR} | definition of parallelogram: line 1 |
| 4. Draw the diagonal \overline{QS} | two points determine a line |
| 5. \angle PQS \cong \angle RSQ | Alternate Interior Angles Theorem: line 2, transversal \overline{QS} |
| 6. \angle PSQ \cong \angle RQS | Alternate Interior Angles Theorem: line 3, transversal \overline{QS} |
| 7. \overline{QS} \cong \overline{SQ} | reflexive property of congruence |
| 8. \triangle PQS \cong \triangle RSQ | ASA: lines 5, 7 and 6 |
| 9. \overline{PQ} \cong \overline{RS} | corresponding parts of congruent triangles: line 8 |
| 10. \overline{PS} \cong \overline{RQ} | corresponding parts of congruent triangles: line 8 |
Line 5 pairs the angle at Q with the angle at S, and line 6 pairs the angle at S with the angle at Q, so the vertices correspond as P \to R, Q \to S, S \to Q. That is the order in line 8. The side \overline{PQ} is the first two letters of the first triangle, so it matches the first two letters of the second, \overline{RS}. The side \overline{PS}, the first and last letters, matches \overline{RQ}. Reading a congruence statement this way is the whole of line 9 and line 10.
A parallelogram with two equal neighbors
In a parallelogram ABCD, \overline{AB} \cong \overline{BC}. Prove that all four sides are congruent, and find the perimeter if AB = 9.
SolutionThis proof is written as a paragraph, which is the same argument as a two-column proof with the lines run together.
By the Opposite Sides Theorem, \overline{CD} \cong \overline{AB} and \overline{DA} \cong \overline{BC}. The Given says \overline{BC} \cong \overline{AB}. So \overline{DA} is congruent to \overline{BC}, and \overline{BC} is congruent to \overline{AB}, which makes \overline{DA} congruent to \overline{AB} by the transitive property of congruence. Now every one of the four sides is congruent to \overline{AB}, so all four are congruent to one another.
If AB = 9, each side is 9 and the perimeter is 4 \cdot 9 = 36.
A parallelogram with four equal sides is a rhombus, and this example showed that one pair of equal neighbors is enough to make it one. More ways to prove a parallelogram is a rhombus are in proving a parallelogram is a rectangle or a rhombus and the shape itself in rectangles, rhombuses and squares.
The other diagonal, and a different reason
In Figure 7, ABCD is a parallelogram. Prove that \angle DAB \cong \angle BCD.
Given. ABCD is a parallelogram.
Prove. \angle DAB \cong \angle BCD.
Plan. The Opposite Sides Theorem is now proved, so it can stand as a reason, and this time it is the tool. It gives two pairs of equal sides, and a third pair is shared, which is enough for SSS. Draw \overline{BD}: then \angle DAB is an angle of \triangle ABD, and \angle BCD is an angle of \triangle CDB, so the angles wanted are parts of the two triangles.
| Statements | Reasons |
|---|---|
| 1. ABCD is a parallelogram | given |
| 2. \overline{AB} \cong \overline{CD} and \overline{AD} \cong \overline{CB} | Opposite Sides Theorem: line 1 |
| 3. Draw the diagonal \overline{BD} | two points determine a line |
| 4. \overline{BD} \cong \overline{DB} | reflexive property of congruence |
| 5. \triangle ABD \cong \triangle CDB | SSS: lines 2 and 4 |
| 6. \angle DAB \cong \angle BCD | corresponding parts of congruent triangles: line 5 |
In line 5 the vertices correspond as A \to C, B \to D, D \to B. Then \overline{AB} matches \overline{CD}, \overline{AD} matches \overline{CB} and \overline{BD} matches \overline{DB}, exactly the three pairs in lines 2 and 4. The angle at A corresponds to the angle at C, which is line 6.
The other pair of opposite angles needs no new work. The triangles ABC and CDA of the first proof in the procedure have B \to D, so \angle ABC \cong \angle CDA by the same congruence. So opposite angles of a parallelogram are congruent, a third property that comes from the first.
The route through vertical angles
Given. ABCD is a parallelogram whose diagonals meet at E, as in Figure 3.
Prove. \overline{AC} and \overline{BD} bisect each other, using the vertical angles at E in place of the second pair of alternate interior angles.
SolutionPlan. The triangles are still \triangle ABE and \triangle CDE. At E the two diagonals cross, so \angle AEB and \angle CED are vertical angles. With one pair of alternate interior angles and the opposite sides, that is two angles and a side.
| Statements | Reasons |
|---|---|
| 1. ABCD is a parallelogram with diagonals meeting at E | given |
| 2. \overline{AB} \parallel \overline{DC} | definition of parallelogram: line 1 |
| 3. \angle BAE \cong \angle DCE | Alternate Interior Angles Theorem: line 2, transversal \overline{AC} |
| 4. \angle AEB \cong \angle CED | Vertical Angles Theorem: \overline{AC} and \overline{BD} cross at E |
| 5. \overline{AB} \cong \overline{CD} | Opposite Sides Theorem: line 1 |
| 6. \triangle ABE \cong \triangle CDE | AAS theorem: lines 3, 4 and 5 |
| 7. \overline{AE} \cong \overline{CE} and \overline{BE} \cong \overline{DE} | corresponding parts of congruent triangles: line 6 |
| 8. \overline{AC} and \overline{BD} bisect each other | definitions of midpoint and segment bisector: line 7, E is on both diagonals |
The side in line 5 is not between the two angles. In \triangle ABE the angles are at A and E, the side between them is \overline{AE}, and \overline{AB} lies opposite the angle at E. So the reason in line 6 is AAS, not ASA. The pairing is the same as before: A \leftrightarrow C, B \leftrightarrow D, E \leftrightarrow E.
From the definition alone
Given. ABCD is a parallelogram whose diagonals meet at E.
Prove. \overline{AC} and \overline{BD} bisect each other, citing nothing about parallelograms except the definition.
SolutionPlan. The proof that the diagonals bisect each other needs \overline{AB} \cong \overline{CD}, and now that has to be proved first. It is a side of \triangle ABC and of \triangle CDA, the two triangles made by the single diagonal \overline{AC} in Figure 2, and those share \overline{AC}. Both pairs of parallel sides give alternate interior angles along \overline{AC}. So this is a two stage proof: one congruence for the side, a second for the halves, which are the triangles of Figure 4.
| Statements | Reasons |
|---|---|
| 1. ABCD is a parallelogram with diagonals meeting at E | given |
| 2. \overline{AB} \parallel \overline{DC} and \overline{AD} \parallel \overline{BC} | definition of parallelogram: line 1 |
| 3. \angle BAC \cong \angle DCA | Alternate Interior Angles Theorem: \overline{AB} \parallel \overline{DC} (line 2), transversal \overline{AC} |
| 4. \angle BCA \cong \angle DAC | Alternate Interior Angles Theorem: \overline{AD} \parallel \overline{BC} (line 2), transversal \overline{AC} |
| 5. \overline{AC} \cong \overline{CA} | reflexive property of congruence |
| 6. \triangle ABC \cong \triangle CDA | ASA postulate: lines 3, 5 and 4 |
| 7. \overline{AB} \cong \overline{CD} | corresponding parts of congruent triangles: line 6 |
| 8. \angle ABD \cong \angle CDB | Alternate Interior Angles Theorem: \overline{AB} \parallel \overline{DC} (line 2), transversal \overline{BD} |
| 9. \triangle ABE \cong \triangle CDE | ASA postulate: lines 3, 7 and 8 |
| 10. \overline{AE} \cong \overline{CE} and \overline{BE} \cong \overline{DE} | corresponding parts of congruent triangles: line 9 |
| 11. \overline{AC} and \overline{BD} bisect each other | definitions of midpoint and segment bisector: line 10, E is on both diagonals |
Line 6 pairs A \leftrightarrow C, B \leftrightarrow D, C \leftrightarrow A, so \overline{AB} goes with \overline{CD}. Line 9 uses line 3 and line 8 for angles of \triangle ABE and \triangle CDE although they were written with other letters. That is allowed because they are the same angles: E lies on \overline{AC}, so the ray \overrightarrow{AE} is the ray \overrightarrow{AC} and \angle BAE is \angle BAC; in the same way \angle DCE is \angle DCA, \angle ABE is \angle ABD and \angle CDE is \angle CDB. Lines 1 to 7 on their own are the proof of the Opposite Sides Theorem, with the same triangles and the same pairing, stopped once \overline{AB} \cong \overline{CD} is reached (the full theorem also reads off \overline{BC} \cong \overline{DA} from line 6). None of it mentions E, which is why citing that theorem in the second proof of the procedure was safe.
A segment through the crossing
Given. ABCD is a parallelogram whose diagonals meet at E. A line through E meets \overline{AB} at P and \overline{CD} at Q, as in Figure 8.
Prove. E is the midpoint of \overline{PQ}.
Plan. \overline{PE} and \overline{QE} are sides of \triangle AEP and \triangle CEQ. The theorem that the diagonals bisect each other, now proved, gives \overline{AE} \cong \overline{CE}. The parallel sides give alternate interior angles at A and C, and \overline{PQ} crossing \overline{AC} at E gives vertical angles. The side \overline{AE} lies between the angles at A and E, so this is ASA.
| Statements | Reasons |
|---|---|
| 1. ABCD is a parallelogram with diagonals meeting at E | given |
| 2. \overline{AE} \cong \overline{CE} | diagonals of a parallelogram bisect each other: line 1 |
| 3. \overline{AB} \parallel \overline{DC} | definition of parallelogram: line 1 |
| 4. \angle PAE \cong \angle QCE | Alternate Interior Angles Theorem: line 3, transversal \overline{AC} |
| 5. \angle AEP \cong \angle CEQ | Vertical Angles Theorem: \overline{PQ} and \overline{AC} cross at E |
| 6. \triangle AEP \cong \triangle CEQ | ASA postulate: lines 4, 2 and 5 |
| 7. \overline{PE} \cong \overline{QE} | corresponding parts of congruent triangles: line 6 |
| 8. E is the midpoint of \overline{PQ} | definition of midpoint: line 7, E is on \overline{PQ} (given) |
P is on \overline{AB}, so \angle PAE is the angle \angle BAC; Q is on \overline{CD}, so \angle QCE is \angle DCA. Those are the alternate interior pair along \overline{AC} from the procedure. The pairing in line 6 is A \leftrightarrow C, E \leftrightarrow E, P \leftrightarrow Q.
Nothing in the proof depended on where the line through E was drawn. Every segment through E with its ends on two opposite sides is cut in half at E, which is the picture of a half turn about E carrying the parallelogram onto itself: the figure has point symmetry about the crossing of its diagonals.
Parallel lines stay the same distance apart
In Figure 9, \ell \parallel m. Points X and Y lie on m, and X' and Y' lie on \ell so that \overline{XX'} \perp \ell and \overline{YY'} \perp \ell. Prove that \overline{XX'} \cong \overline{YY'}.
Given. \ell \parallel m; X and Y are two points of m; \overline{XX'} \perp \ell at X' and \overline{YY'} \perp \ell at Y'.
Prove. \overline{XX'} \cong \overline{YY'}.
Plan. The Prove is about two sides of a quadrilateral, XYY'X'. If that quadrilateral is a parallelogram, the Opposite Sides Theorem finishes the proof. A parallelogram needs two pairs of parallel sides, and one pair is nearly free, because \overline{XY} lies on m and \overline{X'Y'} lies on \ell.
| Statements | Reasons |
|---|---|
| 1. \ell \parallel m | given |
| 2. \overline{XX'} \perp \ell and \overline{YY'} \perp \ell | given |
| 3. \overline{XY} \parallel \overline{X'Y'} | \overline{XY} lies on m and \overline{X'Y'} on \ell: line 1 |
| 4. \angle XX'Y' and \angle YY'Q are right angles, where Q is a point of \ell beyond Y' | definition of perpendicular: line 2 |
| 5. \angle XX'Y' \cong \angle YY'Q | all right angles are congruent: line 4 |
| 6. \overline{XX'} \parallel \overline{YY'} | converse of the Corresponding Angles Theorem: line 5, transversal \ell |
| 7. XYY'X' is a parallelogram | definition of parallelogram: lines 3 and 6 |
| 8. \overline{XX'} \cong \overline{YY'} | Opposite Sides Theorem: line 7 |
Line 6 uses \ell as a transversal of the two perpendiculars. Each perpendicular makes a right angle with \ell, on the same side of \ell as the other and in the matching position at its own foot, so the angles are corresponding and they are congruent. Two lines perpendicular to the same line are therefore parallel. In line 7 the pairs of opposite sides of XYY'X' are \overline{XY} with \overline{Y'X'} and \overline{YY'} with \overline{X'X}, which are the pairs lines 3 and 6 made parallel.
The result is that every point of m is the same distance from \ell. That equal distance is why the distance between parallel lines is one number and not a number that depends on where you measure.
Boundaries
Both pairs of opposite sides must be parallel. Figure 10 has \overline{AB} \parallel \overline{DC} and nothing else, so line 5 of the Opposite Sides proof is available and line 6 is not. The diagonal still makes \angle BAC \cong \angle DCA, but the second pair of angles is missing, ASA cannot run, and the other two sides differ, as the figure shows.
A quadrilateral with one pair of parallel sides is a trapezoid, and in general its other two sides are not congruent. Some trapezoids have congruent legs, but they get them from a different reason, not from this theorem. The diagonals proof fails in a trapezoid too: lines 3 and 4 of it still hold, but line 5 does not. The parallel sides of a trapezoid have different lengths (if they were also equal, the figure would be a parallelogram), and without that pair of sides there is no congruence. The diagonals then cut each other unequally, with E nearer the shorter parallel side.
It is about opposite sides. The neighbors \overline{AB} and \overline{BC} of a parallelogram are congruent only in the special case of Example 7. The theorem says nothing about them.
Each other, not each equal. The diagonals theorem gives AE = CE and BE = DE. It does not give AE = BE: that holds only when the diagonals are the same length, which is what makes a parallelogram a rectangle. Nor do the diagonals of every parallelogram cross at right angles or cut the corner angles in half; those belong to the rhombus. Both special cases are in proving a parallelogram is a rectangle or a rhombus.
The proofs rest on the parallel postulate. The Alternate Interior Angles Theorem, used in both proofs, is itself a consequence of the parallel postulate. So these are theorems about the ordinary flat plane.
The converses are separate theorems. That a quadrilateral with both pairs of opposite sides congruent is a parallelogram is true, but it is not proved by running the Opposite Sides proof backwards. It starts from the sides, gets congruent triangles by SSS, and then needs the converse of the Alternate Interior Angles Theorem. In the same way, a quadrilateral whose diagonals bisect each other is a parallelogram, but that is a separate theorem with its own proof. Both are in proving a quadrilateral is a parallelogram. Citing either theorem of this lesson to show that a shape is a parallelogram argues in a circle, because both start from a parallelogram.
The letters must go around the figure. The diagonals of ABCD are \overline{AC} and \overline{BD}, and its opposite sides are \overline{AB} with \overline{CD}, only because the vertices are named in order. Given three points and asked for “a parallelogram” with no order, there are three possible fourth corners, one for each choice of which two given points are opposite.
A different route exists. A parallelogram is carried onto itself by a half turn about the middle of a diagonal, and a half turn keeps lengths, so each side lands on the one opposite. That proof is short, but it assumes a half turn is a length-preserving motion, which must be established first. The triangle proofs assume only the angle theorems about parallel lines.
Where this goes next
The same diagonals give the rest of the properties of a parallelogram. Opposite angles are congruent, as Example 8 showed, and neighboring angles are supplementary.
Reading the theorems backwards gives ways to show that a quadrilateral is a parallelogram, in the converses: both pairs of opposite sides congruent, or diagonals that bisect each other. From there, adding a condition picks out the special cases: a right angle or congruent diagonals for the rectangle, and a pair of equal neighbors or perpendicular diagonals for the rhombus, in proving a parallelogram is a rectangle or a rhombus.
Example 11 is the start of a different kind of argument. A half turn about E carries the parallelogram onto itself, and proving facts by naming the motion that carries one figure onto another is transformational proof. In coordinates the diagonals theorem becomes one line, since two segments bisect each other exactly when they have the same midpoint, and that line is used again and again in coordinate proofs.
Outside proofs, equal opposite sides are what let a parallelogram be cut and slid into a rectangle of the same area, in why the parallelogram area formula works, and what make adding two arrows by completing a parallelogram give the same answer in either order.
Exercises
In every exercise, ABCD is a parallelogram named in order around it, unless the exercise gives other letters or says otherwise. In 9 to 16, 28 to 31 and 36 to 39 its diagonals meet at E, as in Figure 3.
Opposite sides: using the theorem
- AB = 17 and BC = 9. Find CD, DA and the perimeter.
- In parallelogram PQRS, PQ = 4x + 1 and SR = 7x - 14. Find x and PQ.
- In parallelogram WXYZ, WX = 3n - 2, YZ = n + 10 and XY = 2n + 5. Find n and the perimeter.
- The perimeter is 84 and one side is twice a neighboring side. Find all four sides.
- AB = 2x + 5 and BC = x + 1, and the perimeter is 72. Find x, AB and BC.
- AB = x^2 - 4 and CD = 3x. Find every x that satisfies the equation, and say which make a real parallelogram.
- AB = 2x + y, CD = 13, BC = x + 2y and AD = 8. Find x and y.
- The diagonal \overline{AC} is 10 long, AB = 6 and BC = 8. Find the three sides of \triangle CDA and the perimeter of ABCD.
Diagonals: reading the theorem
- AE = 7 and BE = 11. Find CE, DE, AC and BD.
- AC = 30 and DE = 9. Find AE, BE and BD.
- True or false: in every parallelogram, \overline{AE} \cong \overline{BE}. Explain.
Diagonals: solving for unknowns
- AE = 2x + 5 and CE = 4x - 7. Find x and AC.
- BE = 3y - 2 and BD = 5y + 6. Find y and BD.
- AE = 2a + b, CE = 13, BE = a + 2b and DE = 11. Find a and b.
- AE = x^2 - 5 and CE = 4x. Find every possible value of x, and AC.
- AE = 6 and BE = 8, and in this parallelogram the diagonals happen to be perpendicular. Find AB, BC and the perimeter of ABCD.
Coordinates
- A(-2, 1), B(4, 3), C(6, -1) and D(0, -3) are the vertices of parallelogram ABCD. Find the midpoint of each diagonal.
- A(0, 0), B(5, 1) and C(7, 5) are three vertices of parallelogram ABCD. Find D.
- Parallelogram PQRS has P(1, 2), Q(3, 7) and R(9, 8). Find the point M where its diagonals meet, and the vertex S.
Reading the proofs
- Redo the proof of the Opposite Sides Theorem with the diagonal \overline{BD}. Name the two pairs of alternate interior angles and the two congruent triangles.
- Suppose only \overline{AB} \parallel \overline{DC} is given. Which lines of the ten-line proof of the Opposite Sides Theorem are still supported, and which fall?
- In the proof of the Opposite Sides Theorem, a student writes line 8 as \triangle ABC \cong \triangle ADC. What is wrong with it?
Short proofs
Write each as a two-column proof or a paragraph.
- Given. ABCD is a parallelogram, and E and F lie on the diagonal \overline{AC}, strictly between A and C, with \overline{AE} \cong \overline{CF}. Prove. \overline{BE} \cong \overline{DF}.
- Given. ABCD and ABEF are parallelograms. Prove. \overline{DC} \cong \overline{FE}.
- Given. ABCD is a parallelogram, E is between D and C, F is between A and B, and \overline{DE} \cong \overline{FB}. Prove. \overline{EC} \cong \overline{AF}.
- Given. ABCD is a parallelogram. Prove. Its perimeter is 2(AB + BC).
- Given. ABCD is a parallelogram, M is the midpoint of \overline{AB} and N is the midpoint of \overline{DC}. Prove. \overline{AM} \cong \overline{NC}.
- Prove that the diagonals bisect each other using \triangle ADE and \triangle CBE in place of \triangle ABE and \triangle CDE.
- Prove that \triangle AED \cong \triangle CEB by SSS. You may cite both theorems of this lesson.
- P is on \overline{AE} and Q is on \overline{CE}, with \overline{AP} \cong \overline{CQ}. Prove that E is the midpoint of \overline{PQ}.
- A line through E meets \overline{AD} at P and \overline{BC} at Q. Prove that \overline{PE} \cong \overline{QE}.
Reasoning
- A quadrilateral ABCD has \overline{AB} \parallel \overline{DC}, with AB = 10 and DC = 6. Can it be a parallelogram? Explain.
- Write the converse of the Opposite Sides Theorem. Describe, without writing it out, how a proof of the converse would start and what it would need that this lesson’s proof did not.
- A student says: “Opposite sides of a parallelogram are congruent because parallel lines are always the same distance apart.” Explain why this does not prove the theorem.
- The proof of the Opposite Sides Theorem uses ASA at line 8. Explain why SAS cannot be used there, and what extra step AAS would need.
- The proof that the diagonals bisect each other cites the Opposite Sides Theorem. Explain why that does not make the proof circular.
- ABCD is a trapezoid with \overline{AB} \parallel \overline{DC} and AB \ne DC, and its diagonals meet at E. Which lines of the proof that the diagonals bisect each other still hold, and where does the proof break?
- In kite ABCD, \overline{AB} \cong \overline{AD} and \overline{CB} \cong \overline{CD}, and diagonal \overline{AC} bisects diagonal \overline{BD}. Do the diagonals of a kite bisect each other? Explain, and say what the words “each other” add.
- Explain why the diagonals theorem makes E the center of a half turn that carries the parallelogram onto itself.
Answers
- CD = 17, DA = 9 and the perimeter is 52. Opposite sides are congruent, so 17 + 9 + 17 + 9 = 52.
- x = 5 and PQ = 21. \overline{PQ} and \overline{SR} are opposite, so 4x + 1 = 7x - 14 and 15 = 3x. Then PQ = 4(5) + 1 = 21, and SR = 7(5) - 14 = 21.
- n = 6 and the perimeter is 66. \overline{WX} and \overline{YZ} are opposite, so 3n - 2 = n + 10 and 2n = 12. Then WX = YZ = 16, and XY = ZW = 2(6) + 5 = 17, so the perimeter is 16 + 17 + 16 + 17 = 66.
- The sides are 14, 28, 14 and 28. Call the shorter side s. Opposite sides are congruent, so the perimeter is s + 2s + s + 2s = 6s = 84, and s = 14.
- x = 10, AB = 25 and BC = 11. The perimeter is 2(AB + BC) = 2(3x + 6) = 72, so 3x + 6 = 36 and x = 10.
- x = 4 or x = -1, and only x = 4 makes a parallelogram. x^2 - 4 = 3x gives x^2 - 3x - 4 = 0, which factors as (x - 4)(x + 1) = 0. At x = 4 both sides are 12. At x = -1 both would be -3, and a length cannot be negative.
- x = 6 and y = 1. \overline{AB} is opposite \overline{CD}, so 2x + y = 13, and \overline{BC} is opposite \overline{AD}, so x + 2y = 8. From the second, x = 8 - 2y; putting that into the first gives 16 - 4y + y = 13, so y = 1 and x = 6. Then AB = 13 and BC = 8.
- CD = 6, DA = 8 and AC = 10, so the perimeter of ABCD is 28. \overline{CD} is opposite \overline{AB} and \overline{DA} is opposite \overline{BC}, and \overline{AC} is shared. The perimeter is 2(6 + 8) = 28.
- CE = 7, DE = 11, AC = 14, BD = 22. E is the midpoint of both diagonals, so each half matches its partner and each diagonal is twice its half.
- AE = 15, BE = 9, BD = 18. AE is half of AC; BE matches DE; BD = 2(9).
- False. The theorem pairs each half with the other half of the same diagonal, not with a half of the other diagonal. In Example 4, AE = 14 and BE = 4. The two are equal only when the diagonals are the same length, which happens exactly in a rectangle.
- x = 6 and AC = 34. The halves are equal: 2x + 5 = 4x - 7, so 12 = 2x. Then AE = 2(6) + 5 = 17, and CE = 4(6) - 7 = 17 agrees, so AC = 2(17) = 34.
- y = 10 and BD = 56. BD is twice BE: 5y + 6 = 2(3y - 2) = 6y - 4, so y = 10. Then BD = 5(10) + 6 = 56, and BE = 3(10) - 2 = 28 is half of it.
- a = 5 and b = 3. The bisections give 2a + b = 13 and a + 2b = 11. Doubling the first, 4a + 2b = 26, and subtracting the second leaves 3a = 15. Then b = 13 - 2(5) = 3. Check: 5 + 2(3) = 11.
- x = 5 only, and AC = 40. From x^2 - 5 = 4x, x^2 - 4x - 5 = 0, so (x - 5)(x + 1) = 0. x = -1 gives CE = 4(-1) = -4, and a length cannot be negative, so it is rejected. With x = 5, AE = CE = 20 and AC = 40.
- AB = 10, BC = 10, perimeter 40. \triangle AEB has a right angle at E, so by the Pythagorean theorem AB^2 = 6^2 + 8^2 = 100. CE = AE = 6 by the theorem, and \angle BEC is a right angle because it forms a linear pair with \angle AEB, so BC^2 = 8^2 + 6^2 = 100. Opposite sides are congruent, so the perimeter is 2(10 + 10) = 40.
- Both midpoints are (2, 0). For \overline{AC}: \displaystyle \left(\frac{-2 + 6}{2}, \frac{1 + (-1)}{2}\right) = (2, 0). For \overline{BD}: \displaystyle \left(\frac{4 + 0}{2}, \frac{3 + (-3)}{2}\right) = (2, 0). One point is the middle of both diagonals, as the theorem says it must be.
- D(2, 4). The midpoint of \overline{AC} is \displaystyle \left(\frac{7}{2}, \frac{5}{2}\right), and it is also the midpoint of \overline{BD}: \displaystyle \frac{5 + x}{2} = \frac{7}{2} gives x = 2, and \displaystyle \frac{1 + y}{2} = \frac{5}{2} gives y = 4. Check: A to B and D to C are both 5 right and 1 up.
- M(5, 5) and S(7, 3). The diagonals are \overline{PR} and \overline{QS}, and M is the midpoint of \overline{PR}: \displaystyle \left(\frac{1 + 9}{2}, \frac{2 + 8}{2}\right) = (5, 5). It is also the midpoint of \overline{QS}: \displaystyle \frac{3 + x}{2} = 5 gives x = 7 and \displaystyle \frac{7 + y}{2} = 5 gives y = 3. Check: P to Q and S to R are both 2 right and 5 up.
- The pairs are \angle ABD \cong \angle CDB (from \overline{AB} \parallel \overline{DC}, transversal \overline{BD}) and \angle ADB \cong \angle CBD (from \overline{AD} \parallel \overline{BC}, transversal \overline{BD}). With \overline{BD} shared, ASA gives \triangle ABD \cong \triangle CDB, with A \to C, B \to D, D \to B. Then \overline{AB} \cong \overline{CD} and \overline{AD} \cong \overline{CB}.
- Line 1 falls first: with one pair of parallel sides, ABCD is no longer given to be a parallelogram, so line 2 becomes the Given itself. Lines 2, 4, 5 and 7 stand: the diagonal can still be drawn, line 5 uses only \overline{AB} \parallel \overline{DC} and \overline{AC} is still shared. Line 3 is not given, so line 6 has nothing to rest on, and lines 8, 9 and 10 fall with it: ASA has only one angle pair and one side. Figure 10 shows the loss is real, not just a gap in the proof.
- The letters of \triangle ABC \cong \triangle ADC pair A with A, B with D and C with C. Then \angle BAC would have to match \angle DAC, but lines 5 and 6 match \angle BAC with \angle DCA, the angle at C. The angle at A is paired with the angle at C, so the second triangle must be written \triangle CDA, with C first.
- \triangle ABE \cong \triangle CDF by SAS, so \overline{BE} \cong \overline{DF}. By the Opposite Sides Theorem \overline{AB} \cong \overline{CD}, and by the definition of a parallelogram \overline{AB} \parallel \overline{DC}. Because E and F lie on \overline{AC} between its ends, \angle BAE is the angle \angle BAC and \angle DCF is \angle DCA, a pair of alternate interior angles for transversal \overline{AC}, so \angle BAE \cong \angle DCF. With \overline{AE} \cong \overline{CF} given, each angle lies between the two sides named, and SAS gives \triangle ABE \cong \triangle CDF with A \to C, B \to D, E \to F. Corresponding parts give \overline{BE} \cong \overline{DF}.
- By the Opposite Sides Theorem in ABCD, \overline{DC} \cong \overline{AB}. In ABEF the sides \overline{AB} and \overline{FE} are opposite, so the theorem gives \overline{AB} \cong \overline{FE}. By the transitive property of congruence, \overline{DC} \cong \overline{FE}. The shared side \overline{AB} is what joins the two.
- By the Opposite Sides Theorem, AB = DC. Since F is between A and B, the segment addition postulate gives AB = AF + FB, and since E is between D and C, DC = DE + EC. So AF + FB = DE + EC. The Given says DE = FB, so subtracting that one length from both sides leaves AF = EC, which is \overline{EC} \cong \overline{AF}.
- The perimeter is AB + BC + CD + DA. By the Opposite Sides Theorem CD = AB and DA = BC, so the perimeter is AB + BC + AB + BC = 2(AB + BC).
- By the Opposite Sides Theorem, AB = DC. M is the midpoint of \overline{AB}, so \displaystyle AM = \frac{1}{2}AB, and N is the midpoint of \overline{DC}, so \displaystyle NC = \frac{1}{2}DC. Since AB = DC, half of each is equal, and AM = NC, so \overline{AM} \cong \overline{NC}.
- \overline{AD} \parallel \overline{BC} by the definition of parallelogram. With transversal \overline{AC}, \angle DAE \cong \angle BCE, and with transversal \overline{BD}, \angle ADE \cong \angle CBE, both by the Alternate Interior Angles Theorem. \overline{AD} \cong \overline{CB} by the Opposite Sides Theorem, and it lies between those angles, so \triangle ADE \cong \triangle CBE by ASA, pairing A \leftrightarrow C, D \leftrightarrow B, E \leftrightarrow E. Corresponding parts give \overline{AE} \cong \overline{CE} and \overline{DE} \cong \overline{BE}, so E is the midpoint of both diagonals and they bisect each other.
- \overline{AE} \cong \overline{CE} and \overline{DE} \cong \overline{BE} because the diagonals of a parallelogram bisect each other, and \overline{AD} \cong \overline{CB} by the Opposite Sides Theorem. Three pairs of sides give \triangle AED \cong \triangle CEB by SSS, pairing A \leftrightarrow C, E \leftrightarrow E, D \leftrightarrow B.
- \overline{AE} \cong \overline{CE} because the diagonals bisect each other. P is between A and E, so by the Segment Addition Postulate PE = AE - AP; likewise QE = CE - CQ. Equal lengths minus equal lengths are equal, so PE = QE. P, E and Q all lie on \overline{AC} with E between P and Q, so E is the midpoint of \overline{PQ}.
- \overline{AE} \cong \overline{CE} because the diagonals bisect each other. \overline{AD} \parallel \overline{BC}, so with transversal \overline{AC}, \angle PAE \cong \angle QCE by the Alternate Interior Angles Theorem (\angle PAE is \angle DAC and \angle QCE is \angle BCA). \angle AEP \cong \angle CEQ are vertical angles. So \triangle AEP \cong \triangle CEQ by ASA, with the side between the angles at A and E, and corresponding parts give \overline{PE} \cong \overline{QE}.
- No. If it were a parallelogram, \overline{AB} and \overline{DC} would be opposite sides and so congruent by the theorem, which would make AB = DC. But 10 \ne 6.
- The converse is: if both pairs of opposite sides of a quadrilateral are congruent, then it is a parallelogram. A proof would start from the equal sides, draw a diagonal, and get two triangles congruent by SSS instead of ASA. Corresponding parts then give a pair of alternate interior angles congruent. The last step needs the converse of the Alternate Interior Angles Theorem, to conclude the sides are parallel. This lesson’s proof used the theorem itself, not its converse.
- Equal distance between the parallel lines is about the gap, not about the length of a segment along them. A trapezoid has parallel sides that are the same distance apart everywhere, yet they differ in length. Also, Example 12 proved the equal distance using this theorem, so using it here would be circular.
- SAS needs two pairs of matching sides, but at line 8 only one pair is known, the shared diagonal. The other sides are what is being proved, so using them would be circular. AAS needs two angles and a side not between them, and the diagonal lies between the angles of lines 5 and 6, so AAS cannot use those facts as they stand. It needs one extra step first: the angles of a triangle add to 180^\circ, so the third angles match, \angle ABC \cong \angle CDA. Then \angle ABC, \angle BAC and \overline{AC} give the same congruence by AAS. ASA uses lines 5, 6 and 7 directly, with no extra step.
- The Opposite Sides Theorem is proved from the definition of parallelogram alone: one diagonal \overline{AC}, both pairs of alternate interior angles along it, the shared side, and \triangle ABC \cong \triangle CDA by ASA with A \leftrightarrow C, B \leftrightarrow D, C \leftrightarrow A. It never mentions E or the halves of a diagonal. The first proof in the procedure is that proof, and lines 1 to 7 of Example 10 repeat it, stopped at the pair of sides needed. Since it does not use the diagonals bisecting each other, citing it on the way to that fact is not arguing in a circle.
- Lines 3 and 4 still hold: \overline{AB} \parallel \overline{DC} is now given directly in place of line 2, so both pairs of alternate interior angles are congruent. Line 5 fails, because AB \ne DC, and with it the ASA congruence of line 6 and everything after. The triangles have the same angles but different sizes, and the diagonals do not bisect each other: E divides each one into unequal parts.
- No, not in general. Take A(0, 6), B(-3, 0), C(0, -2) and D(3, 0). Then \displaystyle AB = AD = \sqrt{45} and \displaystyle CB = CD = \sqrt{13}, so ABCD is a kite, and its diagonals meet at (0, 0): \overline{BD} is cut into 3 and 3, but \overline{AC} into 6 and 2. “Each other” demands both at once, each diagonal cutting the other in half. A kite has only one of the two, which is one reason a kite need not be a parallelogram.
- A half turn about E sends each point X to the point X' for which E is the midpoint of \overline{XX'}. E is the midpoint of \overline{AC}, so the turn sends A to C and C to A; E is the midpoint of \overline{BD}, so it sends B to D and D to B. A rotation carries segments to segments, so side \overline{AB} goes to \overline{CD}, \overline{BC} to \overline{DA}, and so on, and the parallelogram lands exactly on itself.