PROBLEM: Hundred Degree Isosceles Triangle

Given an isosceles triangle ABC with AB = AC and the measure of angle BAC = 100 degrees. Extend AC to point D such that AD = BC. Now draw segment BD. What is the size of angle CBD?




Discussion/Solution?:

Construct an equilateral triangle ADE with a side AD and side BE in the interior of angle BAD.
Draw BE.

Each angle of triangle ADE has measure 60, so measure of angle BAE = 40.

We were given AD = BC. We now have AD = DE = EA (by definition of an equilateral triangle). Thus,

BC = EA.

Note also that measure of angle BAE = 40 = measure of angle ACB and that BA = AC (if base angles of a triangle are equal, then the sides opposite those angles are equal). Therefore,

triangle BAC is congruent to triangle EBA.

By corresponding parts of congruent triangles, BE = BA. Also, BD = BD by the reflexive property of equality. So

triangle BAD is congruent to triangle BED

by SSS. This leads to

measure of angle ADB = measure of angle EDB

by corresponding parts. Since measure of angle ADE = 60 , then measure of angle ADB = 30. Thus

measure of angle CBD = 10

(the sum of the angles in a triangle equals 180).

Note the type of angle formed by the intersection of AE and BD.

By labeling the angles in the diagram and using the fact that the sum of the angles in a triangle must equal 180, we find that the angles formed by the intersection of AE and BD are right angles.

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