Problem: In triangle ABC,AC=CD\mathrm{ABC}, \mathrm{AC}=\mathrm{CD}ABC,AC=CD and ∠CAB−∠ABC=30∘\angle \mathrm{CAB}-\angle \mathrm{ABC}=30^{\circ}∠CAB−∠ABC=30∘. Then ∠BAD\angle B A D∠BAD is:
Answer Choices:
A. 30∘30^{\circ}30∘
B. 20∘20^{\circ}20∘
C. 2212∘22 \frac{1}{2}^{\circ}2221∘
D. 10∘10^{\circ}10∘
E. 15∘15^{\circ}15∘ Solution:
∠BAD=∠CAB−∠CAD=∠CAB−∠CDA\angle B A D=\angle C A B-\angle C A D=\angle C A B-\angle C D A ∠BAD=∠CAB−∠CAD=∠CAB−∠CDA
=∠CAB−(∠BAD+∠B). =\angle C A B-(\angle B A D+\angle B) \text {. } =∠CAB−(∠BAD+∠B).
∴2∠BAD=∠CAB−∠B=30∘;∴∠BAD=15∘. \therefore 2 \angle B A D=\angle C A B-\angle B=30^{\circ} ; \quad \therefore \angle B A D=15^{\circ} \text {. } ∴2∠BAD=∠CAB−∠B=30∘;∴∠BAD=15∘.