The isosceles trapezoid

Geometry, grade 9 Lesson 10 of 21Learning

The isosceles trapezoid

The definition of an isosceles trapezoid

Definition: an isosceles trapezoid is a trapezoid whose two legs are equal.

In trapezoid $\displaystyle ABCD$ in the figures below, $\displaystyle AB\parallel DC$ and $\displaystyle {AD}={BC}$ ;
the base angles are the two angles beside the same base: $\displaystyle \angle D$ and $\displaystyle \angle C$ beside the base $\displaystyle DC$ , and $\displaystyle \angle A$ and $\displaystyle \angle B$ beside the base $\displaystyle AB$ .

The base angles of an isosceles trapezoid

\displaystyle {AE}={BF}

because in every trapezoid the heights are equal (the previous lesson). The congruence theorem SSA, from eighth grade: two triangles are congruent if two sides of one equal two sides of the other, and the angles facing the larger of them are equal.

Theorem: in an isosceles trapezoid the base angles are equal.

Given: $\displaystyle AB\parallel DC$ and $\displaystyle {AD}={BC}$ .
Prove: $\displaystyle {\angle D}={\angle C}$ and $\displaystyle {\angle DAB}={\angle CBA}$ .

The proof below drops the heights $\displaystyle AE$ and $\displaystyle BF$ and shows that triangles $\displaystyle ADE$ and $\displaystyle BCF$ are congruent:

	Claim	Reason
1	$\displaystyle {AD}={BC}$	Given
2	$\displaystyle {AE}={BF}$	In a trapezoid all the heights are equal
3	$\displaystyle {\angle AED}={\angle BFC}$	Both are $\displaystyle 90^{\circ}$ ( $\displaystyle AE$ and $\displaystyle BF$ are heights)
4	$\displaystyle \triangle ADE\cong\triangle BCF$	(1), (2), (3); the congruence theorem SSA (the right angle faces the larger side)
5	$\displaystyle {\angle D}={\angle C}$	(4); in congruent triangles, equal sides face equal angles
6	$\displaystyle {\angle DAB}={\angle CBA}$	(5); the angles beside each leg add up to $\displaystyle 180^{\circ}$ , so $\displaystyle {\angle DAB}={180^{\circ}-\angle D}={180^{\circ}-\angle C}={\angle CBA}$

The diagonals of an isosceles trapezoid

Theorem: in an isosceles trapezoid the diagonals are equal.

Given: $\displaystyle AB\parallel DC$ and $\displaystyle {AD}={BC}$ .
Prove: $\displaystyle {AC}={BD}$ .

The diagonals are sides of triangles $\displaystyle ADC$ and $\displaystyle BCD$ . The proof below shows that the two triangles are congruent: they have a leg, a base angle and the base $\displaystyle DC$ , which they share.

	Claim	Reason
1	$\displaystyle {AD}={BC}$	Given
2	$\displaystyle {\angle ADC}={\angle BCD}$	In an isosceles trapezoid the base angles are equal
3	$\displaystyle {DC}={DC}$	A common side
4	$\displaystyle \triangle ADC\cong\triangle BCD$	(1), (2), (3); the congruence theorem SAS
5	$\displaystyle {AC}={BD}$	(4); in congruent triangles, equal angles face equal sides

The converse theorems

To prove that a trapezoid is isosceles, one of three is enough: equal legs, equal base angles or equal diagonals.

The other way round too: a trapezoid with either of these two properties is isosceles.

Theorem: a trapezoid whose base angles are equal is an isosceles trapezoid.
Theorem: a trapezoid whose diagonals are equal is an isosceles trapezoid.

You write both proofs yourself, in Exercise 2 and Exercise 3.

Proving that a quadrilateral is a trapezoid

Two lines that meet at a point are not parallel.

A trapezoid is a quadrilateral with exactly one pair of parallel sides. So there are two things to prove:

one pair of sides is parallel – for example by equal corresponding angles or equal alternate angles;
the other pair is not parallel – for example, because the two sides, extended, meet at a point.

Given: $\displaystyle {AB}={AC}$ .
The point $\displaystyle D$ is on $\displaystyle AB$ , the point $\displaystyle E$ on $\displaystyle AC$ , and $\displaystyle {AD}={AE}$ .
Prove: $\displaystyle DBCE$ is an isosceles trapezoid.

The proof below shows that $\displaystyle DE\parallel BC$ by corresponding angles, and then that the legs $\displaystyle DB$ and $\displaystyle CE$ are equal:

	Claim	Reason
1	$\displaystyle {AB}={AC}$	Given
2	$\displaystyle {AD}={AE}$	Given
3	$\displaystyle {\angle ABC}={\angle ACB}$	(1); in an isosceles triangle the base angles are equal
4	$\displaystyle {\angle ADE}={\angle AED}$	(2); in an isosceles triangle the base angles are equal
5	$\displaystyle {\angle ADE}={\angle ABC}$	(3), (4); triangles $\displaystyle ADE$ and $\displaystyle ABC$ share angle $\displaystyle A$ , and in each of them the other two angles are equal, so each of those is $\displaystyle \frac{180^{\circ}-\angle A}{2}$
6	$\displaystyle DE\parallel BC$	(5); if corresponding angles are equal, the lines are parallel
7	$\displaystyle DBCE$ is a trapezoid	(6); $\displaystyle DB$ and $\displaystyle CE$ are not parallel, since they lie on the lines $\displaystyle AB$ and $\displaystyle AC$ , which meet at $\displaystyle A$
8	$\displaystyle {DB}={CE}$	(1), (2); $\displaystyle {DB}={AB-AD}$ and $\displaystyle {CE}={AC-AE}$
9	$\displaystyle DBCE$ is an isosceles trapezoid	(7), (8); a trapezoid with equal legs is isosceles (the definition)

Exercise 1

$\displaystyle ABCD$ is an isosceles trapezoid, and its bases are $\displaystyle AB$ and $\displaystyle DC$ .

$\displaystyle {\angle DAB}={125^{\circ}}$ .
Find the other angles of the trapezoid, and give the reasons.

\displaystyle \angle ABC

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\displaystyle \angle C

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\displaystyle \angle D

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The perimeter of the trapezoid is $\displaystyle 30$ .
The bases are $\displaystyle {AB}={6}$ and $\displaystyle {DC}={12}$ .
How long is each leg?

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Exercise 2

In trapezoid $\displaystyle ABCD$ the bases are $\displaystyle AB$ and $\displaystyle DC$ , and $\displaystyle {\angle ADC}={\angle BCD}$ .
The segments $\displaystyle AE$ and $\displaystyle BF$ are heights.
Prove: $\displaystyle {AD}={BC}$ , so the trapezoid is isosceles.

Exercise 3

In trapezoid $\displaystyle ABCD$ the bases are $\displaystyle AB$ and $\displaystyle DC$ , and the diagonals are equal: $\displaystyle {AC}={BD}$ .
The segments $\displaystyle AE$ and $\displaystyle BF$ are heights.
Prove: $\displaystyle {AD}={BC}$ , so the trapezoid is isosceles.

Exercise 4

$\displaystyle ABCD$ is an isosceles trapezoid: $\displaystyle AB\parallel DC$ and $\displaystyle {AD}={BC}$ .
The diagonals meet at the point $\displaystyle O$ .
Prove: $\displaystyle {OD}={OC}$ .

Formulas: geometry, grade 9

Open as a page

Angles and sides in a triangle

The sum of the angles of a triangle
$\displaystyle \alpha+\beta+\gamma=180^{\circ}$
An exterior angle $\displaystyle \delta$ equals the sum of the two interior angles not next to it
$\displaystyle \delta$ $\displaystyle =$ $\displaystyle \alpha+\beta$
The larger side faces the larger angle, and the other way round
$\displaystyle a>b\ \Leftrightarrow\ \alpha>\beta$

The right triangle

Pythagoras’ theorem – $\displaystyle a$ , $\displaystyle b$ the legs, $\displaystyle c$ the hypotenuse
$\displaystyle a^{2}+b^{2}$ $\displaystyle =$ $\displaystyle c^{2}$
The leg $\displaystyle a$ facing a $\displaystyle 30^{\circ}$ angle is half the hypotenuse $\displaystyle c$
$\displaystyle a$ $\displaystyle =$ $\displaystyle \frac{1}{2}c$
The median $\displaystyle m_c$ to the hypotenuse is half the hypotenuse $\displaystyle c$
$\displaystyle m_c$ $\displaystyle =$ $\displaystyle \frac{1}{2}c$

Angles between parallel lines

Corresponding angles are equal, and so are alternate angles
$\displaystyle \alpha=\beta$
Co-interior angles
$\displaystyle \alpha+\beta=180^{\circ}$

Angles in a quadrilateral

The sum of the angles of a quadrilateral
$\displaystyle \alpha+\beta+\gamma+\delta=360^{\circ}$
In a trapezoid: the two angles beside a leg
$\displaystyle \alpha+\beta=180^{\circ}$

Areas

The area of a triangle – $\displaystyle h_a$ the height to the side $\displaystyle a$
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{a\cdot h_a}{2}$
The area of a parallelogram – $\displaystyle h_a$ the height to the side $\displaystyle a$
$\displaystyle S$ $\displaystyle =$ $\displaystyle a\cdot h_a$
The area of a rectangle
$\displaystyle S$ $\displaystyle =$ $\displaystyle a\cdot b$
The area of a square
$\displaystyle S$ $\displaystyle =$ $\displaystyle a^{2}$
The area of a trapezoid – $\displaystyle a$ , $\displaystyle b$ the bases, $\displaystyle h$ the height
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{(a+b)\cdot h}{2}$
The area of a rhombus or a kite – $\displaystyle d_1$ , $\displaystyle d_2$ the diagonals
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{d_1\cdot d_2}{2}$
The ratio of the areas of similar triangles – $\displaystyle k$ the ratio of similarity
$\displaystyle \frac{S_1}{S_2}$ $\displaystyle =$ $\displaystyle k^{2}$