Properties of a parallelogram
Two pairs of parallel lines cross and box in a four-sided shape. What does the parallelism alone force to be true about its sides, its angles and its diagonals?
Where you are 7 in Quadrilaterals
Why this exists
Draw two parallel lines, then two more parallel lines across them. Where the four lines cross they box in a four-sided shape, and you chose only two things: two directions. You never asked for any lengths to be equal or any angles to match.
Yet the shape that comes out has equal opposite sides, equal opposite angles, neighboring angles that add to 180^\circ, and diagonals that cut each other exactly in half. That is a great deal of information for free, and all of it comes from the angle facts about a transversal crossing two parallel lines.
It is also why the shape turns up wherever something must move while staying parallel to itself. A folding arm made of four bars, with opposite bars of equal length, is a parallelogram at every position as long as its bars never cross, so as it folds the top bar keeps the direction of the bottom one. This lesson is about what a parallelogram guarantees, so that when a figure says “parallelogram” you know at once which lengths and angles you get for nothing.
Definition
A parallelogram is a quadrilateral whose two pairs of opposite sides are parallel.
The vertices are named in order around the shape. In parallelogram ABCD the sides are \overline{AB}, \overline{BC}, \overline{CD} and \overline{DA}, and the definition says
Arrowheads, as in Figure 1, are the usual way to mark this: sides carrying the same number of arrowheads are parallel.
A few names for the parts are used all the time.
- Opposite sides share no vertex: \overline{AB} and \overline{CD}, or \overline{BC} and \overline{DA}.
- Opposite angles share no side: \angle A and \angle C, or \angle B and \angle D.
- Consecutive angles are neighbors that share a side, such as \angle A and \angle B, which share \overline{AB}.
- The diagonals join opposite vertices: \overline{AC} and \overline{BD}.
Writing \angle A for \angle DAB is safe while only one angle sits at A. Once a diagonal is drawn, three angles meet at A, and the full three-letter name is needed.
Notice what the definition leaves out. It says nothing about lengths, nothing about the sizes of the angles and nothing about the diagonals. Every result below is forced by the two pairs of parallel sides alone.
Intuition
Two pictures make the results believable before they are proved.
A slide. Side \overline{DC} is side \overline{AB} slid along \overline{AD} without turning. A slide changes where a segment is and nothing else, so \overline{DC} is exactly as long as \overline{AB}. In the same way, \overline{BC} is \overline{AD} slid along \overline{AB}.
A half-turn. Put a pin where the diagonals cross, at E in Figure 2, and turn the whole parallelogram through 180^\circ about it. It lands exactly on itself, with A and C trading places and B and D trading places. A turn moves a shape without stretching it, so whatever trades places must match: \overline{AB} lands on \overline{CD}, \angle A lands on \angle C, and \overline{EA} lands on \overline{EC}. A figure that a half-turn carries onto itself has point symmetry, and E is its center.
Neither picture is a proof. The half-turn, for instance, takes for granted that the shape lands on itself, which is the very thing to be shown. The argument in “Why it is true” uses nothing but parallel lines, angles and triangles.
Results
Let ABCD be a parallelogram whose diagonals meet at E.
Opposite sides are congruent.
Opposite angles are congruent.
Consecutive angles are supplementary.
and the same holds for \angle B and \angle C, for \angle C and \angle D, and for \angle D and \angle A.
The diagonals bisect each other. Each diagonal cuts the other into two equal halves, so E is the midpoint of both.
In Figure 3 the ticks mark equal sides and the arcs mark equal angles. One measured angle settles all four: if m\angle A = a^\circ, then going around the angles are a^\circ, (180 - a)^\circ, a^\circ and (180 - a)^\circ. They add to 360^\circ, as the angles of any quadrilateral must.
Notice what is not on the list. In general the diagonals are not equal, they do not cross at right angles, and a diagonal does not cut the angles at its ends in half. Those are extra facts about special parallelograms, and the Boundaries section comes back to them.
Why it is true
Every result comes from one fact about a transversal crossing two parallel lines, used first on the sides and then on the diagonals.
Consecutive angles. In Figure 4 the side \overline{AB}, extended, is a transversal crossing the parallel lines AD and BC. The angles \angle 1 and \angle 2 lie between those two parallels and on the same side of the transversal, so they are same side interior angles, and same side interior angles between parallel lines are supplementary:
Opposite angles. Now take \overline{BC}, extended, as the transversal, crossing the parallel lines AB and DC. The same fact gives
Both sums equal 180^\circ, so they equal each other. Subtracting m\angle 2 from both sides leaves m\angle 1 = m\angle 3. This is the argument that proves vertical angles equal, with same side interior angles in place of linear pairs. The pair \angle B and \angle D works the same way.
Opposite sides. Draw the diagonal \overline{AC}. It is a transversal for both pairs of parallel sides. Because \overline{AB} \parallel \overline{DC}, the alternate interior angles \angle BAC and \angle DCA are congruent. Because \overline{AD} \parallel \overline{BC}, the alternate interior angles \angle BCA and \angle DAC are congruent. The side \overline{AC} belongs to both triangles and lies between those angles. Two angles and the side between them match, so \triangle ABC \cong \triangle CDA by the ASA criterion, and matching parts of congruent triangles are equal: AB = CD and BC = DA.
Diagonals. Draw both diagonals, meeting at E. In \triangle ABE and \triangle CDE, the sides \overline{AB} and \overline{CD} are congruent, as just shown. With \overline{AC} as a transversal of the parallels AB and DC, the alternate interior angles \angle BAE and \angle DCE are congruent; with \overline{BD} as the transversal, so are \angle ABE and \angle CDE. So \triangle ABE \cong \triangle CDE by ASA, which gives AE = CE and BE = DE.
Each of these arguments is set out line by line, with every reason cited, in proving properties of a parallelogram.
Worked examples
All four angles from one
In parallelogram ABCD, m\angle A = 64^\circ. Find the other three angles.
Solution\angle C is opposite \angle A, and opposite angles of a parallelogram are congruent, so m\angle C = 64^\circ.
\angle B is consecutive to \angle A: they share the side \overline{AB}. Consecutive angles are supplementary, so
\angle D is opposite \angle B, so m\angle D = 116^\circ.
Check. 64 + 116 + 64 + 116 = 360, as the angles of a quadrilateral must add to.
Opposite angles written in x
In Figure 5, ABCD is a parallelogram with m\angle A = (3x + 15)^\circ and m\angle C = (5x - 21)^\circ. Find x and all four angles.
\angle A and \angle C are opposite angles, so they are equal. Set the expressions equal and solve:
Substitute into both expressions, since they must agree: 3(18) + 15 = 69 and 5(18) - 21 = 69. So m\angle A = m\angle C = 69^\circ.
\angle B and \angle D are each consecutive to \angle A, so each measures 180^\circ - 69^\circ = 111^\circ.
The angles are 69^\circ, 111^\circ, 69^\circ and 111^\circ.
Consecutive angles written in y
In parallelogram ABCD, m\angle A = (2y + 10)^\circ and m\angle B = (3y - 5)^\circ. Find y and all four angles.
SolutionThis time the two angles are neighbors, not opposite, so they are not equal. They are consecutive angles, so they add to 180^\circ:
Then m\angle A = 2(35) + 10 = 80^\circ and m\angle B = 3(35) - 5 = 100^\circ, and 80 + 100 = 180 as it should be. The opposite angles copy these: m\angle C = 80^\circ and m\angle D = 100^\circ.
Whether two angles are opposite or consecutive decides the equation. Opposite means equal; consecutive means adding to 180^\circ. Read that off the letters before writing anything.
A diagonal and the angles it makes
In Figure 6, ABCD is a parallelogram with diagonal \overline{AC}, and m\angle BAC = 32^\circ and m\angle CAD = 45^\circ. Find m\angle ACD, m\angle BCA and all four angles of the parallelogram.
The diagonal \overline{AC} is a transversal for both pairs of parallel sides, so it carries angles from A across to C.
\overline{AB} \parallel \overline{DC}, and \angle BAC and \angle ACD are alternate interior angles for that pair, so they are equal: m\angle ACD = 32^\circ.
\overline{AD} \parallel \overline{BC}, and \angle CAD and \angle BCA are alternate interior angles for that pair, so m\angle BCA = 45^\circ.
The ray AC lies inside \angle DAB, so by angle addition m\angle DAB = 32^\circ + 45^\circ = 77^\circ. In the same way m\angle BCD = 45^\circ + 32^\circ = 77^\circ, which agrees with opposite angles being equal. Each consecutive angle is the supplement:
Check. The angles of \triangle ABC are 32^\circ, 103^\circ and 45^\circ, which add to 180^\circ, as the angles of a triangle must.
The diagonal did not cut \angle A in half: one part is 32^\circ and the other 45^\circ. Nothing about a parallelogram forces it to.
Triangles made by the diagonals
The diagonals of parallelogram ABCD meet at E. AC = 18, BD = 24 and AB = 13. Find the perimeter of \triangle CDE.
SolutionThe diagonals bisect each other, so E is the midpoint of each:
Opposite sides are congruent, so CD = AB = 13. The perimeter of \triangle CDE is 9 + 12 + 13 = 34.
Check. \triangle ABE has sides AE = 9, BE = 12 and AB = 13: the same three lengths. That is no coincidence, since \triangle ABE \cong \triangle CDE is exactly how the bisecting was proved. The other two triangles, \triangle BCE and \triangle DAE, match each other as well, but nothing here says they match the first pair, because BC was never given.
Checking every property on a grid
The points A(-1, 2), B(5, 4), C(7, 9) and D(1, 7) are joined in order. Show that ABCD is a parallelogram, then check that its opposite sides are equal and its diagonals bisect each other. Are its diagonals equal?
SolutionParallel sides. Two lines are parallel when they have equal slopes and are not the same line.
Line DC is not line AB, since D lies far above line AB, and likewise BC is not AD. Both pairs of opposite sides are parallel, so by the definition ABCD is a parallelogram.
Opposite sides. From A to B the run is 6 and the rise 2, and from D to C it is the same. From A to D the run is 2 and the rise 5, and from B to C it is the same. So by the distance formula
Diagonals bisect. Find the midpoint of each diagonal:
The two midpoints are the same point, so the diagonals cross there and each one is cut in half.
Diagonal lengths.
These are not equal, since \displaystyle \sqrt{113} is more than 10. The diagonals bisect each other without being equal. Equal diagonals belong to rectangles, not to parallelograms in general.
The bisectors of two consecutive angles
In parallelogram ABCD, the bisector of \angle A and the bisector of \angle B meet at P. Show that m\angle APB = 90^\circ, whatever the shape of the parallelogram.
SolutionNo measure is given, so name one. Let m\angle A = a^\circ. Consecutive angles are supplementary, so m\angle B = (180 - a)^\circ.
A bisector cuts an angle into two equal halves, so in \triangle ABP
These two add to \displaystyle \frac{a + 180 - a}{2} = 90 degrees. The three angles of \triangle ABP add to 180^\circ, so
The value of a dropped out, so the two bisectors meet at a right angle in every parallelogram. The supplementary pair is what makes it work: halve two angles that add to 180^\circ and the halves add to 90^\circ.
A bisector that cuts off an isosceles triangle
In Figure 7, ABCD is a parallelogram with AB = 10 and AD = 6. The bisector of \angle A meets side \overline{DC} at E. Find DE and EC.
The bisector splits \angle A into two equal parts, so \angle DAE \cong \angle EAB.
\overline{AB} \parallel \overline{DC} and \overline{AE} is a transversal, so the alternate interior angles \angle EAB and \angle AED are congruent.
Put the two facts together: \angle DAE \cong \angle AED. Those are two angles of \triangle ADE, and by the converse of the base angles theorem a triangle with two equal angles has equal sides opposite them. The side opposite \angle AED is \overline{AD}, and the side opposite \angle DAE is \overline{DE}, so
Opposite sides of the parallelogram are congruent, so DC = AB = 10, and
Two parallelogram facts did the work, the parallel sides (through the alternate interior angles) and the equal opposite sides, together with one triangle fact. The answer also shows when E lands on the side at all: here AD is shorter than AB, so DE = AD fits inside \overline{DC}. Were AD longer than AB, the bisector would cross \overline{BC} instead and meet line DC only beyond C.
Boundaries
The results start from a parallelogram. Each one is about a quadrilateral already known to have both pairs of opposite sides parallel. Each also works backwards: a quadrilateral with both pairs of opposite sides congruent, or both pairs of opposite angles congruent, or diagonals that bisect each other, must be a parallelogram. Those converses are separate theorems, proved in proving a quadrilateral is a parallelogram.
Half a condition is not enough. A quadrilateral with only one pair of parallel sides is a trapezoid, and its other two sides can have any lengths. One pair of congruent opposite sides is not enough either: an isosceles trapezoid has its two legs equal and is not a parallelogram. A kite that is not a rhombus has one pair of equal opposite angles and is not one either.
The diagonals bisect each other, and that is all. In a general parallelogram the diagonals have different lengths (Example 6), they do not cross at right angles, and they do not bisect the angles at their ends (Example 4). Equal diagonals happen exactly when the parallelogram is a rectangle, and perpendicular diagonals exactly when it is a rhombus; both belong to rectangles, rhombuses and squares.
Neighbors need not match. Opposite parts are equal. Consecutive angles add to 180^\circ and are equal only when every angle is a right angle, and two neighboring sides can have any two lengths.
Four real corners. The four vertices must not have three on one line. Points placed along two parallel lines with three of them in a row do not make a quadrilateral, and none of the results apply.
The letters go in order. ABCD names the vertices in order around the shape. If they are listed out of order, \overline{AC} may be a side rather than a diagonal, and “opposite” means something else.
Where this goes next
The properties here are tools; proving them carefully is the next step. Proving properties of a parallelogram writes out the congruent-triangle arguments as full proofs, and proofs about parallelograms gathers that whole family of proofs together. The converses, which let you decide that a shape is a parallelogram, are proving a quadrilateral is a parallelogram, and on a grid they become classifying figures from coordinates, as in Example 6.
Add one more condition and a parallelogram becomes special: a right angle makes it a rectangle, and two equal neighboring sides make it a rhombus. Each special shape keeps every property in this lesson and gains more of its own. What each special quadrilateral guarantees and the quadrilateral family tree sort out which shape inherits what.
The slide picture from the Intuition section becomes a formula too. Cut a triangle off one end of a parallelogram and slide it to the other end, and you have a rectangle with the same base and the same height, which is the area of a parallelogram.
Exercises
In every exercise, ABCD is a parallelogram named in order around it, unless the exercise says otherwise. Where its diagonals are mentioned, they meet at E.
Angles from one fact
- m\angle A = 47^\circ. Find m\angle B, m\angle C and m\angle D.
- m\angle D = 90^\circ. Find the other three angles.
- m\angle B is four times m\angle A. Find all four angles.
- m\angle C is 50^\circ less than m\angle D. Find all four angles.
Angles in algebra
- m\angle A = (6x - 8)^\circ and m\angle C = (4x + 22)^\circ. Find x and all four angles.
- m\angle B = (7y + 3)^\circ and m\angle C = (5y - 3)^\circ. Find y and all four angles.
- m\angle A = (x^2 - 20)^\circ and m\angle C = (x + 22)^\circ. Find every value of x that works, and m\angle A for each.
- m\angle A = (2a + b)^\circ, m\angle B = (a + 3b)^\circ and m\angle C = 70^\circ. Find a and b.
Sides and diagonals together
- AE = 5, BE = 7 and AB = 9. Find the perimeter of \triangle CDE.
- AC = 16, BD = 22 and BC = 12. Find the perimeter of \triangle BCE.
- The diagonal \overline{AC} is 15 long, and the perimeter of \triangle ABC is 40. Find the perimeter of ABCD.
A diagonal or a bisector
- m\angle BAC = 40^\circ and m\angle DAC = 25^\circ. Find m\angle BCA, m\angle DCA, m\angle BCD and m\angle ABC.
- m\angle ABC = 110^\circ and m\angle ABD = 70^\circ. Find m\angle ADB, m\angle BDC and m\angle DAB.
- AD = 7 and DC = 12. The bisector of \angle A meets \overline{DC} at F. Find DF and FC.
Could it be a parallelogram?
Each exercise describes a quadrilateral, named in order around it. Decide whether it could be a parallelogram, and give a reason.
- m\angle A = 70^\circ, m\angle B = 110^\circ, m\angle C = 75^\circ and m\angle D = 105^\circ.
- AB = 8, BC = 5, CD = 8 and DA = 6.
- The diagonals of ABCD meet at E, with AE = CE = 6, BE = 4 and DE = 5.
- PQRS with P(0, 0), Q(6, 1), R(8, 5) and S(2, 4).
- KLMN with K(1, 1), L(5, 1), M(6, 4) and N(1, 4).
Reasoning
- One angle of a parallelogram is a right angle. Explain why all four are.
- Can a parallelogram have three acute angles? Explain.
- True or false: a diagonal of a parallelogram always bisects the two angles at its ends. Explain.
- True or false: the two diagonals of a parallelogram always have the same length. Explain.
- In parallelogram ABCD with AB > AD, the bisector of \angle A meets \overline{DC} at F and the bisector of \angle C meets \overline{AB} at G. Prove that \overline{AF} \parallel \overline{GC}.
- A quadrilateral ABCD has m\angle A + m\angle B = 180^\circ and m\angle B + m\angle C = 180^\circ. Prove that it is a parallelogram.
Answers
- m\angle B = 133^\circ, m\angle C = 47^\circ, m\angle D = 133^\circ. \angle C is opposite \angle A; \angle B is consecutive to it, so 180^\circ - 47^\circ = 133^\circ; \angle D is opposite \angle B.
- All four are 90^\circ. \angle B is opposite \angle D, and \angle A and \angle C are each consecutive to it, so each is 180^\circ - 90^\circ = 90^\circ. The parallelogram is a rectangle.
- m\angle A = m\angle C = 36^\circ and m\angle B = m\angle D = 144^\circ. Consecutive angles add to 180^\circ: a + 4a = 180, so 5a = 180 and a = 36.
- m\angle A = m\angle C = 65^\circ and m\angle B = m\angle D = 115^\circ. \angle C and \angle D are consecutive: d + (d - 50) = 180, so 2d = 230 and d = 115.
- x = 15; m\angle A = m\angle C = 82^\circ and m\angle B = m\angle D = 98^\circ. Opposite angles are equal: 6x - 8 = 4x + 22, so 2x = 30. Check: 6(15) - 8 = 82 = 4(15) + 22.
- y = 15; m\angle B = m\angle D = 108^\circ and m\angle A = m\angle C = 72^\circ. Consecutive angles: (7y + 3) + (5y - 3) = 180, so 12y = 180. Then 7(15) + 3 = 108 and 5(15) - 3 = 72.
- x = 7, giving m\angle A = 29^\circ, or x = -6, giving m\angle A = 16^\circ. Opposite angles: x^2 - 20 = x + 22, so x^2 - x - 42 = 0 and (x - 7)(x + 6) = 0. Both work: 49 - 20 = 29 = 7 + 22 and 36 - 20 = 16 = -6 + 22, and each angle is between 0^\circ and 180^\circ. A negative x is fine; only the angle must be positive.
- a = 20 and b = 30. Opposite angles: 2a + b = 70. Consecutive angles: a + 3b = 180 - 70 = 110. From the first, b = 70 - 2a; then a + 210 - 6a = 110, so 5a = 100. Check: 2(20) + 30 = 70 and 20 + 3(30) = 110.
- 21. The diagonals bisect each other, so CE = AE = 5 and DE = BE = 7; opposite sides give CD = AB = 9. Then 5 + 7 + 9 = 21.
- 31. BE = \tfrac{1}{2}(22) = 11 and CE = \tfrac{1}{2}(16) = 8, so the perimeter is 11 + 8 + 12 = 31.
- 50. AB + BC = 40 - 15 = 25. Opposite sides are equal, so the perimeter of ABCD is 2(AB + BC) = 2(25) = 50.
- m\angle BCA = 25^\circ, m\angle DCA = 40^\circ, m\angle BCD = 65^\circ and m\angle ABC = 115^\circ. With \overline{AD} \parallel \overline{BC}, \angle BCA and \angle DAC are alternate interior angles; with \overline{AB} \parallel \overline{DC}, so are \angle DCA and \angle BAC. Then 25 + 40 = 65, and \angle ABC is consecutive to \angle BCD: 180 - 65 = 115. Check in \triangle ABC: 40 + 115 + 25 = 180.
- m\angle ADB = 40^\circ, m\angle BDC = 70^\circ and m\angle DAB = 70^\circ. First m\angle CBD = 110 - 70 = 40. With \overline{AD} \parallel \overline{BC} and transversal \overline{BD}, \angle ADB \cong \angle CBD. With \overline{AB} \parallel \overline{DC}, \angle BDC \cong \angle ABD. \angle DAB is consecutive to \angle ABC: 180 - 110 = 70. Check in \triangle ABD: 70 + 70 + 40 = 180.
- DF = 7 and FC = 5. As in Example 8, \angle DAF \cong \angle FAB (the bisector) and \angle FAB \cong \angle AFD (alternate interior angles), so \triangle ADF has two equal angles and DF = AD = 7. Then FC = 12 - 7 = 5.
- No. \angle A and \angle C are opposite angles, and 70^\circ and 75^\circ differ. The angles do add to 360^\circ, and since 70 + 110 = 180 the sides \overline{AD} and \overline{BC} are parallel, so the shape can exist, as a trapezoid.
- No. \overline{BC} and \overline{DA} are opposite sides, and 5 and 6 differ.
- No. E is the midpoint of \overline{AC} but not of \overline{BD}, since 4 and 5 differ, and the diagonals of a parallelogram bisect each other.
- Yes. The slopes of PQ and SR are both \displaystyle \frac{1}{6}, and the slopes of QR and PS are both \displaystyle \frac{4}{2} = 2, so both pairs of opposite sides are parallel. As a check, both diagonals have midpoint (4, 2.5).
- No. KL and NM both have slope 0, so they are parallel, but KN is vertical while LM has slope 3, so the other pair is not parallel. KLMN is a trapezoid; also KL = 4 while NM = 5.
- Consecutive angles add to 180^\circ, so each neighbor of the right angle measures 180^\circ - 90^\circ = 90^\circ, and the opposite angle equals it, 90^\circ. All four are right angles, so the parallelogram is a rectangle.
- No. If \angle A is acute, its two neighbors \angle B and \angle D each measure 180^\circ minus an acute angle, so they are obtuse. At most two angles are acute, and they are a pair of opposite angles.
- False. In Example 4 the diagonal \overline{AC} splits \angle A into 32^\circ and 45^\circ. It does happen in a rhombus, but not in a parallelogram in general.
- False. In Example 6 the diagonals measure \displaystyle \sqrt{113} and 5. A parallelogram has equal diagonals only when it is a rectangle.
- Since AB > AD, the argument of Example 8 puts F strictly between D and C, and the same argument at C gives BG = BC = AD < AB, so G is strictly between A and B. The bisector gives m\angle BAF = \tfrac{1}{2} m\angle A. At G, \angle BGC and \angle GCD are alternate interior angles for \overline{AB} \parallel \overline{DC}, so m\angle BGC = m\angle GCD = \tfrac{1}{2} m\angle C, and m\angle C = m\angle A because they are opposite angles. So \angle BAF \cong \angle BGC. These are corresponding angles where the transversal AB crosses lines AF and GC: both open toward B, and F and C lie on the same side of AB. Equal corresponding angles make the lines parallel, so \overline{AF} \parallel \overline{GC}.
- Side \overline{AB} is a transversal of lines AD and BC, and \angle A and \angle B are same side interior angles for it. They add to 180^\circ, so by the converse of the same side interior angles theorem, \overline{AD} \parallel \overline{BC}. In the same way, with \overline{BC} as the transversal of lines AB and DC, m\angle B + m\angle C = 180^\circ gives \overline{AB} \parallel \overline{DC}. Both pairs of opposite sides are parallel, so ABCD is a parallelogram by the definition.