Problem:
Let △ABC be an isosceles triangle with BC=AC and ∠ACB=40∘ Construct the circle with diameter BC, and let D and E be the other intersection points of the circle with the sides AC and AB, respectively. Let F be the intersection of the diagonals of the quadrilateral BCDE. What is the degree measure of ∠BFC?
Answer Choices:
A. 90
B. 100
C. 105
D. 110
E. 120
Solution:
Because BC=AC and ∠ACB=40∘, it follows that ∠BAC=∠ABC=70∘. Because ∠BAC=21​(BC−DE) and BC=180∘, it follows that DE=40∘. Then
Because D and E lie on the circle with diameter BC, both ∠BDC and ∠BEC are right angles, so ∠ADF and ∠AEF are also right angles. Therefore in quadrilateral AEFD
