Question

Triangles $ABC$ and $ADC$ are isosceles with $AB = BC$ and $AD = DC.$ Point $D$ is inside $\triangle ABC,$ $\angle ABC = 40^\circ,$ and $\angle ADC = 140^\circ.$ What is the degree measure of $\angle BAD?$

$20$

$30$

$40$

$50$

$60$

Answer: D

Concepts:isosceles triangle angle chasing

Difficulty rating: 1170

Small Hint:

In an isosceles triangle the two base angles are equal

Big Hint:

Find $\angle BAC$ and $\angle DAC,$ then subtract

Solution:

Since $\triangle ABC$ is isosceles, $\angle BAC = \tfrac12(180^\circ - 40^\circ) = 70^\circ.$

Since $\triangle ADC$ is isosceles, $\angle DAC = \tfrac12(180^\circ - 140^\circ) = 20^\circ.$

Therefore $\angle BAD = \angle BAC - \angle DAC$ $= 70^\circ - 20^\circ$ $= 50^\circ.$

Thus, the correct answer is D.

2007 AMC 10A Problem 8

Answer: D

Difficulty rating: 1170

Small Hint:

Big Hint:

Solution:

Problem 8 in Other Years