Question

triangles acd and bcd are isosceles. angle bac has a measure of 33° and angle bdc has a measure of 35°. find the measure of angle abd. type your answer in the box. angle abd has a measure of °.

Explanation:

Step1: Find angle \( \angle ACD \) in \( \triangle ACD \)

Since \( AD = AC \) in \( \triangle ACD \), and \( \angle BAC=33^{\circ} \), then \( \angle ADC=\angle ACD \). Using the angle - sum property of a triangle (\( \angle A+\angle ADC+\angle ACD = 180^{\circ} \)), and \( \angle A = 33^{\circ} \), we have \( 2\angle ACD=180^{\circ}-\angle BAC \). So \( \angle ACD=\frac{180 - 33}{2}=73.5^{\circ} \).

Step2: Find angle \( \angle BCD \) in \( \triangle BCD \)

Since \( BD = BC \) in \( \triangle BCD \), and \( \angle BDC = 35^{\circ} \), then \( \angle BCD=\angle BDC = 35^{\circ} \) (base - angles of an isosceles triangle are equal).

Step3: Use the exterior - angle property

\( \angle ABD\) is an exterior angle of \( \triangle BCD \) and \( \triangle ABC \). By the exterior - angle property of a triangle (\( \angle ABD=\angle BCD+\angle BAC+\angle BDC \)). Substitute \( \angle BAC = 33^{\circ} \) and \( \angle BCD=\angle BDC = 35^{\circ} \). So \( \angle ABD=33^{\circ}+35^{\circ}+35^{\circ} \)

Answer:

\(103\)