Problem: In △ABC,∠A=55∘,∠C=75∘,D\triangle A B C, \angle A=55^{\circ}, \angle C=75^{\circ}, D△ABC,∠A=55∘,∠C=75∘,D is on side AB‾\overline{A B}AB and EEE is on side BC‾\overline{B C}BC. If DB=BED B=B EDB=BE, then ∠BED=\angle B E D=∠BED=
Answer Choices:
A. 50∘50^{\circ}50∘
B. 55∘55^{\circ}55∘
C. 60∘60^{\circ}60∘
D. 65∘65^{\circ}65∘
E. 70∘70^{\circ}70∘
Solution:
We have ∠B=180∘−(55∘+75∘)=50∘\angle B=180^{\circ}-\left(55^{\circ}+75^{\circ}\right)=50^{\circ}∠B=180∘−(55∘+75∘)=50∘. Since △BDE\triangle B D E△BDE is isosceles, ∠BED=180∘−∠B2=180∘−50∘2=65∘\angle B E D=\dfrac{180^{\circ}-\angle B}{2}=\dfrac{180^{\circ}-50^{\circ}}{2}=65^{\circ}∠BED=2180∘−∠B=2180∘−50∘=65∘.
