Proving a quadrilateral is a parallelogram
A parallelogram is defined by two pairs of parallel sides, and parallel is hard to see. Which easier facts about the sides, the angles or the diagonals are enough to guarantee it?
Where you are 4 in Proofs about parallelograms
When you reach for it
A proof or a problem needs a quadrilateral to be a parallelogram, and nobody has said that it is. The definition asks for both pairs of opposite sides to be parallel, and parallel is a hard thing to show head on. Lengths, angles and midpoints are much easier to get.
The lessons before this one went the other way. Starting from a parallelogram, they proved that opposite sides are congruent and that the diagonals bisect each other. This lesson turns those facts round: which of them, on its own, is enough to force a parallelogram? Each answer is the converse of a theorem already proved, and a converse is a new statement that needs its own proof, since a true theorem can have a false converse (a theorem and its converse).
Four tests work, and the definition itself makes a fifth:
| Test | What must be shown about ABCD |
|---|---|
| definition | \overline{AB} \parallel \overline{DC} and \overline{AD} \parallel \overline{BC} |
| Opposite Sides Converse | \overline{AB} \cong \overline{CD} and \overline{BC} \cong \overline{DA} |
| Opposite Angles Converse | \angle A \cong \angle C and \angle B \cong \angle D |
| Diagonals Converse | \overline{AC} and \overline{BD} bisect each other |
| One Pair Test | \overline{AB} \parallel \overline{DC} and \overline{AB} \cong \overline{DC}, the same pair |
Half of any row is not enough, and the Boundaries show a shape for each half that fails.
Throughout, ABCD is a quadrilateral named in order around it: four segments \overline{AB}, \overline{BC}, \overline{CD} and \overline{DA} that meet only at their ends, with no two neighboring sides along one line. It may be convex or not; each test below holds for both. So \overline{AB} is opposite \overline{CD}, \overline{BC} is opposite \overline{DA}, and the diagonals are \overline{AC} and \overline{BD}. The angle \angle A means the angle of the quadrilateral at A, which is \angle DAB.
The procedure
The target is always the definition. However the proof starts, it has to end with two pairs of parallel sides, or with a test already proved to lead there. The way to show two lines parallel is to find an angle pair and use one of the angle theorems backwards, in proving lines parallel: if alternate interior angles are congruent the lines are parallel, and if same-side interior angles are supplementary the lines are parallel. So each proof below is a route from what is given to a pair of angles.
The Opposite Sides Converse. If both pairs of opposite sides of a quadrilateral are congruent, the quadrilateral is a parallelogram.
Given. \overline{AB} \cong \overline{CD} and \overline{BC} \cong \overline{DA}.
Prove. ABCD is a parallelogram.
The Given is about lengths and the target is about angles. Congruent triangles turn one into the other, so draw the diagonal \overline{AC}: it cuts the quadrilateral into \triangle ABC and \triangle CDA, and all three pairs of sides are then known.
| Statements | Reasons |
|---|---|
| 1. \overline{AB} \cong \overline{CD} and \overline{BC} \cong \overline{DA} | given |
| 2. Draw the diagonal \overline{AC} | two points determine a line |
| 3. \overline{AC} \cong \overline{CA} | reflexive property of congruence |
| 4. \triangle ABC \cong \triangle CDA | SSS: lines 1 and 3 |
| 5. \angle BAC \cong \angle DCA and \angle BCA \cong \angle DAC | corresponding parts of congruent triangles: line 4 |
| 6. \overline{AB} \parallel \overline{DC} | converse of the Alternate Interior Angles Theorem: line 5, transversal \overline{AC} |
| 7. \overline{AD} \parallel \overline{BC} | converse of the Alternate Interior Angles Theorem: line 5, transversal \overline{AC} |
| 8. ABCD is a parallelogram | definition of parallelogram: lines 6 and 7 |
In line 4 the vertices correspond as A \to C, B \to D, C \to A, so \overline{AB} matches \overline{CD}, \overline{BC} matches \overline{DA} and \overline{AC} matches \overline{CA}: exactly the pairs of lines 1 and 3 (congruence statements and correspondence). The same correspondence gives line 5. The angle at A in \triangle ABC, between \overline{AB} and the diagonal, matches the angle at C in \triangle CDA, between \overline{CD} and the diagonal.
In line 6, \angle BAC and \angle DCA are alternate interior angles for the lines \overleftrightarrow{AB} and \overleftrightarrow{DC} because B and D lie on opposite sides of the diagonal, a position fact the figure shows (what a diagram may assume). The Boundaries explain why it always holds in a quadrilateral with these sides. Line 7 is the same move with the other angle pair and the lines \overleftrightarrow{AD} and \overleftrightarrow{BC}.
This is the proof that the lesson on opposite sides said the converse would need: SSS in place of ASA, and the converse of the angle theorem in place of the theorem itself.
The Opposite Angles Converse. If both pairs of opposite angles of a quadrilateral are congruent, the quadrilateral is a parallelogram.
Given. \angle A \cong \angle C and \angle B \cong \angle D.
Prove. ABCD is a parallelogram.
No diagonal is needed. The four angles of a quadrilateral add to 360^\circ (angles in a quadrilateral), and with m\angle A = m\angle C = x and m\angle B = m\angle D = y that sum says
So every angle is supplementary to both of its neighbors, and supplementary neighbors are same-side interior angles.
| Statements | Reasons |
|---|---|
| 1. \angle A \cong \angle C and \angle B \cong \angle D | given |
| 2. m\angle A + m\angle B + m\angle C + m\angle D = 360^\circ | the angles of a quadrilateral add to 360^\circ |
| 3. 2\,m\angle A + 2\,m\angle B = 360^\circ, so m\angle A + m\angle B = 180^\circ | substitution: lines 1 and 2, then division by 2 |
| 4. m\angle A + m\angle D = 180^\circ | substitution: m\angle D = m\angle B (line 1) into line 3 |
| 5. \overline{AD} \parallel \overline{BC} | converse of the Same-Side Interior Angles Theorem: line 3, transversal \overline{AB} |
| 6. \overline{AB} \parallel \overline{DC} | converse of the Same-Side Interior Angles Theorem: line 4, transversal \overline{AD} |
| 7. ABCD is a parallelogram | definition of parallelogram: lines 5 and 6 |
In line 5 the side \overline{AB} crosses the lines \overleftrightarrow{AD} and \overleftrightarrow{BC}, and \angle DAB and \angle ABC lie between them, on the side of \overline{AB} where C and D are. Line 6 is the same with the side \overline{AD} crossing \overleftrightarrow{AB} and \overleftrightarrow{DC}. That C and D lie on one side of \overleftrightarrow{AB} is safe here: x and y are positive and add to 180^\circ, so every angle is less than 180^\circ, and a quadrilateral like that is convex.
The Diagonals Converse. If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram.
Given. The diagonals \overline{AC} and \overline{BD} meet at E, with \overline{AE} \cong \overline{CE} and \overline{BE} \cong \overline{DE}.
Prove. ABCD is a parallelogram.
The forward proof went from parallel sides to congruent triangles to equal halves. This one runs the other way. The halves and the vertical angles at E give two sides and the angle between them, which is SAS, and the triangles then give alternate interior angles.
| Statements | Reasons |
|---|---|
| 1. \overline{AC} and \overline{BD} meet at E; \overline{AE} \cong \overline{CE} and \overline{BE} \cong \overline{DE} | given |
| 2. \angle AEB \cong \angle CED | Vertical Angles Theorem |
| 3. \triangle AEB \cong \triangle CED | SAS: lines 1, 2 and 1 |
| 4. \angle EAB \cong \angle ECD | corresponding parts of congruent triangles: line 3 |
| 5. \overline{AB} \parallel \overline{DC} | converse of the Alternate Interior Angles Theorem: line 4, transversal \overline{AC} |
| 6. \angle AED \cong \angle CEB | Vertical Angles Theorem |
| 7. \triangle AED \cong \triangle CEB | SAS: lines 1, 6 and 1 |
| 8. \angle EAD \cong \angle ECB | corresponding parts of congruent triangles: line 7 |
| 9. \overline{AD} \parallel \overline{BC} | converse of the Alternate Interior Angles Theorem: line 8, transversal \overline{AC} |
| 10. ABCD is a parallelogram | definition of parallelogram: lines 5 and 9 |
In line 3 the angle at E lies between \overline{EA} and \overline{EB} in one triangle and between \overline{EC} and \overline{ED} in the other, so the correspondence is A \to C, E \to E, B \to D, and the angle at A matches the angle at C. E lies on \overline{AC}, so \angle EAB is the angle \angle CAB and \angle ECD is \angle ACD: a pair of alternate interior angles along the diagonal.
This time the position fact is proved rather than read from the figure. E lies on the segment \overline{BD}, between B and D, and also on the line \overleftrightarrow{AC}. A segment that meets a line at a point between its ends has its ends on opposite sides of that line, so B and D are on opposite sides of \overleftrightarrow{AC}, and lines 5 and 9 are about genuine alternate interior angles.
The One Pair Test. If one pair of opposite sides of a quadrilateral is both congruent and parallel, the quadrilateral is a parallelogram.
Given. \overline{AB} \parallel \overline{DC} and \overline{AB} \cong \overline{DC}.
Prove. ABCD is a parallelogram.
One pair of sides is already parallel, so only the other pair is left. Draw the diagonal \overline{AC}. The parallel sides give one pair of alternate interior angles along it, the congruent sides give one pair of sides, and the diagonal is shared: SAS.
| Statements | Reasons |
|---|---|
| 1. \overline{AB} \parallel \overline{DC} and \overline{AB} \cong \overline{DC} | given |
| 2. Draw the diagonal \overline{AC} | two points determine a line |
| 3. \angle BAC \cong \angle DCA | Alternate Interior Angles Theorem: line 1, transversal \overline{AC} |
| 4. \overline{AC} \cong \overline{CA} | reflexive property of congruence |
| 5. \triangle BAC \cong \triangle DCA | SAS: lines 1, 3 and 4 |
| 6. \angle BCA \cong \angle DAC | corresponding parts of congruent triangles: line 5 |
| 7. \overline{AD} \parallel \overline{BC} | converse of the Alternate Interior Angles Theorem: line 6, transversal \overline{AC} |
| 8. ABCD is a parallelogram | definition of parallelogram: lines 1 and 7 |
In line 5 the angle at A lies between \overline{AB} and \overline{AC}, and the angle at C between \overline{CD} and \overline{CA}, so each is the angle between the two sides named, and the correspondence is B \to D, A \to C, C \to A. Under it the angle at C in \triangle BAC matches the angle at A in \triangle DCA, which is line 6.
Lines 3 and 7 both need B and D on opposite sides of \overleftrightarrow{AC}. That always holds here. If B and D were on the same side, then \overrightarrow{AB} and \overrightarrow{DC} would point opposite ways along the two parallel lines, and the sides \overline{BC} and \overline{DA}, each running from one line to the other, would cross, as in the last figure under Boundaries.
This proof uses both directions of one theorem: line 3 is the Alternate Interior Angles Theorem, from parallel to congruent, and line 7 is its converse, from congruent to parallel.
Choosing a test. Let the information decide. Four lengths point to the Opposite Sides Converse and four angles to the Opposite Angles Converse. Facts about a crossing point point to the Diagonals Converse. A pair of sides known to be parallel, such as pieces of the sides of a larger parallelogram, points to the One Pair Test. In coordinates the Diagonals Converse is often the shortest, since it is two midpoints and one comparison. Once proved, each test can stand as the reason for a single line in a later proof, the way the Opposite Sides Theorem does.
Worked examples
Which facts are enough
Each of (a) to (d) gives facts about a quadrilateral ABCD. Decide whether they guarantee a parallelogram, and name the test, or a shape that has the facts and is not a parallelogram.
(a) AB = 7, BC = 4, CD = 7 and DA = 4.
(b) m\angle A = 70^\circ, m\angle B = 110^\circ and m\angle C = 70^\circ.
(c) \overline{AB} \parallel \overline{DC} and AD = BC = 5.
(d) \overline{AB} \parallel \overline{DC} and AB = DC = 9.
Solution(a) Yes. \overline{AB} and \overline{CD} are opposite, and both are 7 long; \overline{BC} and \overline{DA} are opposite, and both are 4 long. Both pairs of opposite sides are congruent, so the Opposite Sides Converse applies.
(b) Yes, although only three angles are given. The four angles add to 360^\circ, so the fourth is forced:
Now \angle A \cong \angle C and \angle B \cong \angle D, and the Opposite Angles Converse applies.
(c) No. An isosceles trapezoid, like the one drawn under Boundaries, has one pair of sides parallel and the other two sides equal, and its parallel sides have different lengths. The parallel pair and the congruent pair here are different pairs.
(d) Yes. The same pair, \overline{AB} and \overline{DC}, is both parallel and congruent, which is the One Pair Test. Parts (c) and (d) differ only in which pair carries the lengths, and that difference is the whole of the test.
Unknowns that make the angles match
In quadrilateral ABCD, m\angle A = (4x + 6)^\circ, m\angle B = (3y - 6)^\circ, m\angle C = (6x - 24)^\circ and m\angle D = (2y + 34)^\circ. Find the values of x and y that make ABCD a parallelogram.
SolutionTwo theorems do different jobs here. Opposite angles of a parallelogram are congruent (proved in the lesson on opposite sides), so if ABCD is a parallelogram, \angle A \cong \angle C and \angle B \cong \angle D: no other values can work. The Opposite Angles Converse says that when both pairs are congruent, ABCD is a parallelogram: these values do work.
From \angle A \cong \angle C:
so m\angle A = m\angle C = 4(15) + 6 = 66^\circ. From \angle B \cong \angle D, 3y - 6 = 2y + 34, so y = 40 and m\angle B = m\angle D = 3(40) - 6 = 114^\circ.
Check. m\angle C = 6(15) - 24 = 66^\circ and m\angle D = 2(40) + 34 = 114^\circ. The four angles add to 66 + 114 + 66 + 114 = 360 degrees, as a quadrilateral’s must. That check is not a formality: the expressions came from a problem, not a drawing, and if the sum had not been 360^\circ no quadrilateral would have these angles at all. Neighboring angles add to 66 + 114 = 180 degrees, as a parallelogram’s do.
Two tests on one set of points
The points are A(-1, 0), B(4, 1), C(6, 5) and D(1, 4). Show that ABCD is a parallelogram, first by the One Pair Test and then by the Diagonals Converse.
SolutionThe One Pair Test on \overline{AB} and \overline{DC}. Parallel lines have equal slopes:
Equal slopes mean the lines are parallel or are one line. On \overleftrightarrow{AB}, the point with x = 1 is 2 steps right of A, so it has \displaystyle y = \frac{1}{5} \cdot 2 = \frac{2}{5}, not 4, so D is not on it and the lines are different. So \overline{AB} \parallel \overline{DC}. The lengths come from the distance formula:
One pair of opposite sides is both parallel and congruent, so ABCD is a parallelogram.
The Diagonals Converse. The midpoint of \overline{AC} is \displaystyle \left(\frac{-1 + 6}{2}, \frac{0 + 5}{2}\right) = \left(\frac{5}{2}, \frac{5}{2}\right), and the midpoint of \overline{BD} is \displaystyle \left(\frac{4 + 1}{2}, \frac{1 + 4}{2}\right) = \left(\frac{5}{2}, \frac{5}{2}\right). One point is the midpoint of both diagonals, so the diagonals meet there and cut each other in half, and ABCD is a parallelogram.
The two tests agree, as they must.
A parallelogram inside a parallelogram
Given. ABCD is a parallelogram. E lies on \overline{AB} and F on \overline{CD}, each strictly between the ends of its side, with \overline{BE} \cong \overline{DF}.
Prove. AECF is a parallelogram.
SolutionPlan. Named in order, AECF has sides \overline{AE}, \overline{EC}, \overline{CF} and \overline{FA}, so \overline{AE} and \overline{CF} are opposite. Both lie along sides of ABCD, which are parallel, so the One Pair Test needs only their lengths, and those are what is left of two equal sides after two equal pieces are taken off.
| Statements | Reasons |
|---|---|
| 1. ABCD is a parallelogram; E is between A and B, F between C and D; \overline{BE} \cong \overline{DF} | given |
| 2. AB = CD | Opposite Sides Theorem: line 1 |
| 3. AE = AB - EB and CF = CD - FD | Segment Addition Postulate: line 1 |
| 4. AE = CF | subtraction: lines 2, 3 and 1 |
| 5. \overline{AB} \parallel \overline{DC} | definition of parallelogram: line 1 |
| 6. \overline{AE} \parallel \overline{CF} | \overline{AE} lies on \overleftrightarrow{AB} and \overline{CF} on \overleftrightarrow{DC}: line 5 |
| 7. AECF is a parallelogram | One Pair Test: lines 4 and 6 |
Line 2 uses the forward theorem, which is allowed because ABCD is given to be a parallelogram. Line 7 uses a converse, which is needed because AECF is not. Both kinds of theorem sit in one proof, each on the figure it is about.
One pair of parallel sides and one pair of opposite angles
Given. In quadrilateral ABCD, \overline{AB} \parallel \overline{DC} and \angle A \cong \angle C.
Prove. ABCD is a parallelogram.
SolutionPlan. None of the four tests fits as it stands, so go back to the definition. One pair of sides is parallel already. For the other pair, find two supplementary same-side interior angles along \overline{DC}. The given parallel sides make \angle A and \angle D supplementary, and \angle C can stand in for \angle A.
| Statements | Reasons |
|---|---|
| 1. \overline{AB} \parallel \overline{DC} and \angle A \cong \angle C | given |
| 2. m\angle A + m\angle D = 180^\circ | Same-Side Interior Angles Theorem: line 1, transversal \overline{AD} |
| 3. m\angle C + m\angle D = 180^\circ | substitution: m\angle C = m\angle A (line 1) into line 2 |
| 4. \overline{AD} \parallel \overline{BC} | converse of the Same-Side Interior Angles Theorem: line 3, transversal \overline{DC} |
| 5. ABCD is a parallelogram | definition of parallelogram: lines 1 and 4 |
In line 2, \angle DAB and \angle ADC lie between the parallel lines \overleftrightarrow{AB} and \overleftrightarrow{DC}, on the same side of \overline{AD}. In line 4, \angle BCD and \angle CDA lie between \overleftrightarrow{CB} and \overleftrightarrow{DA}, on the same side of \overline{DC}, so their being supplementary makes those lines parallel.
The angles must be opposite ones. With \overline{AB} \parallel \overline{DC} and \angle A \cong \angle B instead, the isosceles trapezoid under Boundaries fits the facts, and it is not a parallelogram.
A linkage that keeps a tray level
A tool box has a tray that swings out on two arms. The frame bar \overline{AB} is fixed and level. The arms \overline{AD} and \overline{BC} are both 25 cm long, the tray bar \overline{DC} is exactly as long as \overline{AB}, and all four bars are joined by pins that let them turn. Explain why the tray stays level in every position, as long as the bars form a quadrilateral.
SolutionThe angles of ABCD change as the arms swing, so the machine fixes no angle. The lengths do not change: the bars are rigid. At every position, DC = AB and AD = BC = 25 cm, so both pairs of opposite sides are congruent. By the Opposite Sides Converse, ABCD is a parallelogram in every position, so \overline{DC} \parallel \overline{AB}. The frame bar is level, so the tray bar is level too.
The side test fits because its hypothesis is exactly the part of the machine that never moves. It also explains the condition in the question. If the arms are folded flat, all four bars lying along one line, the linkage can come out of that position with two bars crossing, like the figure at the end of the Boundaries, and then the tray tilts. The theorem is about quadrilaterals, and the crossed shape is not one.
Boundaries
Half a test proves nothing. Every test above needs both of its parts, and each half on its own is shared by a shape that is not a parallelogram.
- One pair of opposite angles. The kite ABCD has \angle B \cong \angle D, but its other two angles, at A and at C, differ.
- One diagonal cutting the other in half. In the same kite, \overline{AC} cuts \overline{BD} in half, but \overline{BD} does not cut \overline{AC} in half. “Each other” means both at once.
- Two pairs of congruent sides that are neighbors. The kite has \overline{AB} \cong \overline{AD} and \overline{CB} \cong \overline{CD}. The Opposite Sides Converse needs the congruent sides to be opposite; in a kite they meet at a corner.
- One pair of opposite sides congruent, or one pair parallel and the other pair congruent. The isosceles trapezoid below has \overline{AB} \parallel \overline{DC} and \overline{AD} \cong \overline{BC}. The One Pair Test asks for one pair to be both, and here the parallel pair and the congruent pair are different pairs.
A shape that is not convex never passes a test. The dart PQRS below is a kite pushed in at R, so its angle inside at R is more than 180^\circ and it is not convex. Like the kite, it has two pairs of congruent neighboring sides and \angle Q \cong \angle S, yet it is far from a parallelogram.
No whole test can be met by a shape like the dart. For the Opposite Angles Converse, the angle opposite the one over 180^\circ would have to be over 180^\circ as well, and the four angles would add to more than 360^\circ. For the Diagonals Converse, a quadrilateral is convex exactly when its two diagonals, as segments, cross each other. In the dart, \overline{PR} stops at R and \overline{QS} lies outside the shape, so they never meet and cannot cut each other in half. The other two tests were proved with no convexity assumed, and they end in a parallelogram, which is convex. So a dart can satisfy half of a test, but never a whole one.
The four sides must meet only at their ends. In the figure below, A, B, C and D are joined in order with \overline{AB} \cong \overline{CD} and \overline{BC} \cong \overline{DA}, yet no side is parallel to the side opposite it. The sides \overline{BC} and \overline{DA} cross, so this is not a quadrilateral, and the Opposite Sides Converse says nothing about it.
The proof shows where it breaks. SSS still gives \triangle ABC \cong \triangle CDA, but B and D now lie on the same side of \overleftrightarrow{AC}, so \angle BAC and \angle DCA are not alternate interior angles and line 6 has no reason. This is also why the position fact in that proof always holds for a real quadrilateral. If B and D were on the same side of \overleftrightarrow{AC}, the congruence would make D the mirror image of B across the perpendicular bisector of \overline{AC}, and then two opposite sides would cross, as they do here.
Each converse starts from a quadrilateral not known to be a parallelogram. So none of the parallelogram properties may be cited inside its proof. Saying that opposite sides of a parallelogram are congruent tells nothing about a figure until it is known to be one.
The flat plane. The Opposite Angles Converse uses the angle sum of a quadrilateral, and the One Pair Test uses the Alternate Interior Angles Theorem, and both of those rest on the parallel postulate, so those two proofs are proofs about the ordinary flat plane. The proofs of the Opposite Sides and Diagonals Converses use only triangle congruence, vertical angles and the converse of the angle theorem, and that converse is proved without the postulate.
Where this goes next
A parallelogram is where the special quadrilaterals start. Once a figure passes one of these tests, one more condition picks out a rectangle, in proving a parallelogram is a rectangle, or a rhombus, in proving a parallelogram is a rhombus, and the family tree of all of them is in the quadrilateral hierarchy.
In coordinates the tests become calculations: equal midpoints of the diagonals, or equal slopes and equal lengths for one pair of sides, as in Example 3. That is how a shape on a grid is named in classifying a quadrilateral on the plane and how general facts are proved in coordinate proofs.
The One Pair Test has a short form with arrows. A pair of opposite sides both parallel and equal, and pointing the same way round, is the statement that the arrow from D to C is the arrow from A to B, which is where adding two arrows by completing a parallelogram gets its shape. And joining the midpoints of the sides of any quadrilateral at all, convex or not, gives a parallelogram, a fact proved with the One Pair Test and the triangle midsegment.
Exercises
In every exercise, ABCD is a quadrilateral named in order around it, unless the exercise says otherwise.
Which facts are enough
For each of 1 to 7, decide whether the facts guarantee that ABCD is a parallelogram. If they do, name the test. If not, name a shape that has the facts and is not a parallelogram, or explain why ABCD cannot be one.
- AB = 8, BC = 5, CD = 8 and DA = 5.
- m\angle A = 75^\circ, m\angle B = 105^\circ and m\angle C = 75^\circ.
- \overline{AB} \parallel \overline{DC} and AD = BC.
- The diagonals meet at E, with AE = CE = 6, BE = 4 and DE = 9.
- \overline{AB} \parallel \overline{DC} and AB = DC = 12.
- AB = AD = 5 and CB = CD = 9.
- \angle A \cong \angle C, and nothing else is known.
Solving for unknowns
- AB = 3x - 4, CD = x + 10, BC = 2y and DA = y + 6. Find the x and y that make ABCD a parallelogram by the Opposite Sides Converse, and find AB and BC.
- The diagonals meet at E, with AE = 5a - 3, CE = 2a + 9, BE = b + 7 and DE = 3b - 5. Find the a and b that make the diagonals bisect each other, and find AC and BD.
- m\angle A = (5x - 12)^\circ, m\angle B = (2y + 10)^\circ, m\angle C = (3x + 28)^\circ and m\angle D = (3y - 31)^\circ. Find the x and y that make ABCD a parallelogram, and all four angles.
- \overline{AB} \parallel \overline{DC}, AB = 4x + 1 and DC = 6x - 11. Find the x that makes ABCD a parallelogram, and find AB.
- m\angle A = (2x + 10)^\circ, m\angle C = (3x - 20)^\circ and m\angle B = m\angle D = 70^\circ. Find the x that makes \angle A \cong \angle C, and say whether ABCD is then a parallelogram.
Coordinates
- A(0, 0), B(5, 2), C(7, 7) and D(2, 5). Use the Diagonals Converse to show that ABCD is a parallelogram.
- P(-3, 1), Q(2, -1), R(5, 3) and S(0, 5). Use the One Pair Test on \overline{PQ} and \overline{SR} to show that PQRS is a parallelogram.
- K(1, 1), L(6, 1), M(5, 4) and N(2, 4). Show that \overline{KL} \parallel \overline{NM} and KN = LM, and then show that KLMN is not a parallelogram.
Proofs
Write each as a two-column proof or a paragraph.
- Given. In quadrilateral ABCD, \angle A is supplementary to \angle B and to \angle D. Prove. ABCD is a parallelogram.
- Given. ABCD and ABEF are parallelograms, and DCEF, named in that order, is a quadrilateral. Prove. DCEF is a parallelogram.
- Given. In quadrilateral ABCD, \overline{AB} \cong \overline{CD} and \angle BAC \cong \angle DCA. Prove. ABCD is a parallelogram.
- Given. The diagonals of quadrilateral ABCD meet at E, and \triangle AEB \cong \triangle CED. Prove. ABCD is a parallelogram.
Reasoning
- In the proof of the Diagonals Converse, it was proved, not read from the figure, that B and D lie on opposite sides of \overleftrightarrow{AC}. Explain how.
- A quadrilateral has one pair of opposite angles congruent, and one of its diagonals cuts the other in half. Must it be a parallelogram? Explain.
- Explain why a quadrilateral that is not convex can never pass the Opposite Angles Converse or the Diagonals Converse.
- Write the Opposite Sides Theorem and the Opposite Sides Converse together as one “if and only if” statement. Explain why the proof of one direction cannot serve as the proof of the other.
- Four sticks, two 10 long and two 6 long, are joined end to end by hinges to make a quadrilateral. Explain why it is a parallelogram in every position when the equal sticks are opposite each other, and describe the shape it makes when the equal sticks are next to each other.
Answers
- Yes, by the Opposite Sides Converse. \overline{AB} and \overline{CD} are opposite and both 8; \overline{BC} and \overline{DA} are opposite and both 5.
- Yes, by the Opposite Angles Converse. The angles add to 360^\circ, so m\angle D = 360^\circ - 75^\circ - 105^\circ - 75^\circ = 105^\circ. Then \angle A \cong \angle C at 75^\circ and \angle B \cong \angle D at 105^\circ.
- Not necessarily. An isosceles trapezoid has \overline{AB} \parallel \overline{DC} and AD = BC, with AB \ne DC. The parallel pair and the equal pair are different pairs, so the One Pair Test does not apply.
- It cannot be a parallelogram. E is the midpoint of \overline{AC}, but BE = 4 and DE = 9, so E is not the midpoint of \overline{BD}. The diagonals of a parallelogram bisect each other, and these do not.
- Yes, by the One Pair Test. The same pair, \overline{AB} and \overline{DC}, is both parallel and congruent.
- It cannot be. The equal sides are neighbors, so ABCD is a kite. A parallelogram would need the opposite sides \overline{AB} and \overline{CD} to be equal, and 5 \ne 9.
- Not necessarily. A kite with AB = BC and CD = DA is symmetric about the diagonal \overline{BD}, so \angle A \cong \angle C. When AB \ne DA it is not a parallelogram, because its opposite sides \overline{AB} and \overline{CD} differ.
- x = 7, y = 6, AB = 17 and BC = 12. From 3x - 4 = x + 10, 2x = 14, and then AB = CD = 17. From 2y = y + 6, y = 6, and then BC = DA = 12. Both pairs of opposite sides are congruent.
- a = 4, b = 6, AC = 34 and BD = 26. From 5a - 3 = 2a + 9, 3a = 12, and then AE = CE = 17. From b + 7 = 3b - 5, 12 = 2b, and then BE = DE = 13. Each diagonal is twice its half.
- x = 20 and y = 41; m\angle A = m\angle C = 88^\circ and m\angle B = m\angle D = 92^\circ. From 5x - 12 = 3x + 28, 2x = 40. From 2y + 10 = 3y - 31, y = 41. Check: the angles add to 88 + 92 + 88 + 92 = 360 degrees.
- x = 6 and AB = 25. The sides are already parallel, so the One Pair Test needs only AB = DC: 4x + 1 = 6x - 11 gives 12 = 2x. Then AB = 25 and DC = 6(6) - 11 = 25.
- x = 30 makes m\angle A = m\angle C = 70^\circ, but then all four angles are 70^\circ and they add to 280^\circ, not 360^\circ. No quadrilateral has these angles, so there is no ABCD to be a parallelogram: the facts are impossible.
- The midpoint of \overline{AC} is \displaystyle \left(\frac{0 + 7}{2}, \frac{0 + 7}{2}\right) = \left(\frac{7}{2}, \frac{7}{2}\right), and the midpoint of \overline{BD} is \displaystyle \left(\frac{5 + 2}{2}, \frac{2 + 5}{2}\right) = \left(\frac{7}{2}, \frac{7}{2}\right). One point is the midpoint of both diagonals, so they bisect each other and ABCD is a parallelogram.
- The slope of \overline{PQ} is \displaystyle \frac{-1 - 1}{2 - (-3)} = -\frac{2}{5} and the slope of \overline{SR} is \displaystyle \frac{3 - 5}{5 - 0} = -\frac{2}{5}. S is not on \overleftrightarrow{PQ}, since that line crosses x = 0 at y = -1/5, not 5, so \overline{PQ} \parallel \overline{SR}. Also \displaystyle PQ = \sqrt{5^2 + 2^2} = \sqrt{29} and \displaystyle SR = \sqrt{5^2 + 2^2} = \sqrt{29}. One pair is both parallel and congruent.
- \overline{KL} and \overline{NM} both have slope 0, so they are parallel, and \displaystyle KN = \sqrt{1^2 + 3^2} = \sqrt{10} and \displaystyle LM = \sqrt{(-1)^2 + 3^2} = \sqrt{10}. But the midpoint of \overline{KM} is \displaystyle \left(3, \frac{5}{2}\right) and the midpoint of \overline{LN} is \displaystyle \left(4, \frac{5}{2}\right). The diagonals of a parallelogram share a midpoint, so KLMN is not one. It is an isosceles trapezoid, with KL = 5 and NM = 3.
- First, ABCD is convex: m\angle B = m\angle D = 180^\circ - m\angle A, so the angle sum leaves m\angle C = m\angle A, and all four angles are under 180^\circ. So \angle DAB and \angle ABC are same-side interior angles for the lines \overleftrightarrow{AD} and \overleftrightarrow{BC} with transversal \overline{AB}, and they are supplementary, so \overline{AD} \parallel \overline{BC} by the converse of the Same-Side Interior Angles Theorem. \angle DAB and \angle ADC are same-side interior angles for \overleftrightarrow{AB} and \overleftrightarrow{DC} with transversal \overline{AD}, so in the same way \overline{AB} \parallel \overline{DC}. Both pairs of opposite sides are parallel, so ABCD is a parallelogram by definition.
- In ABCD, \overline{DC} \cong \overline{AB} by the Opposite Sides Theorem and \overline{DC} \parallel \overline{AB} by the definition. In ABEF the sides \overline{AB} and \overline{FE} are opposite, so \overline{AB} \cong \overline{FE} and \overline{AB} \parallel \overline{FE}. By the transitive property, \overline{DC} \cong \overline{FE}, and two lines parallel to the same line are parallel, so \overline{DC} \parallel \overline{FE} (they are not one line, since DCEF is a quadrilateral). In DCEF the sides \overline{DC} and \overline{EF} are opposite, so by the One Pair Test DCEF is a parallelogram.
- \angle BAC and \angle DCA are alternate interior angles for \overleftrightarrow{AB} and \overleftrightarrow{DC} with transversal \overline{AC}, and they are congruent, so \overline{AB} \parallel \overline{DC} by the converse of the Alternate Interior Angles Theorem. With \overline{AB} \cong \overline{CD} given, the same pair is both parallel and congruent, and the One Pair Test finishes the proof. (The angles are alternate interior because B and D are on opposite sides of \overleftrightarrow{AC}. On one side, SAS would make \triangle BAC \cong \triangle DCA with D the mirror image of B across the perpendicular bisector of \overline{AC}, and two sides would cross, as in the Boundaries.)
- The congruence \triangle AEB \cong \triangle CED pairs A \to C, E \to E, B \to D, so corresponding parts give \overline{AE} \cong \overline{CE} and \overline{BE} \cong \overline{DE}. E lies on both diagonals, so it is the midpoint of each, the diagonals bisect each other, and the Diagonals Converse makes ABCD a parallelogram.
- E is where the diagonals meet, so it lies on the segment \overline{BD}, strictly between B and D, and it also lies on \overleftrightarrow{AC}. The segment from B to D crosses the line \overleftrightarrow{AC} at a point between its ends, and a segment does that only when its ends are on opposite sides of the line.
- Not necessarily. The kite ABCD in the Boundaries has \angle B \cong \angle D, and its diagonal \overline{AC} cuts \overline{BD} in half, yet it is not a parallelogram, since AB \ne CD. The tests need both pairs of opposite angles, or both diagonals cut in half.
- A quadrilateral that is not convex has one angle greater than 180^\circ. The Opposite Angles Converse would need the angle opposite it to be the same size, and two angles over 180^\circ already add to more than 360^\circ. Its diagonals, as segments, do not meet (one lies outside the shape, and the other stops at the corner pushed in), so they cannot cut each other in half, and the Diagonals Converse cannot be met.
- A quadrilateral is a parallelogram if and only if both pairs of its opposite sides are congruent. The forward proof starts from parallel sides, uses the Alternate Interior Angles Theorem and ASA, and ends with lengths. The converse starts from lengths, uses SSS and the converse of the Alternate Interior Angles Theorem, and ends with parallel sides. Running a proof backwards would need every step to reverse, and the steps are different theorems, each of which needed its own proof.
- With the equal sticks opposite, AB = CD = 10 and BC = DA = 6 in every position, because the sticks do not stretch, so the Opposite Sides Converse makes it a parallelogram each time (as long as the sticks do not cross). With the equal sticks next to each other, say AB = BC = 10 and CD = DA = 6, it is a kite, or a dart if it is pushed in, and never a parallelogram, since the opposite sides \overline{AB} and \overline{CD} differ.