imc

Question

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 ?

Accepted Answer

FIGURE Drawing it out, we see BDC and BEC are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find ABC=70^. We can find ECB=20^ and DBC=50^ by the triangle angle sum on ECB and DBC. BDC+ DCB+ DBC=180^90^+40^+ DBC=180^ DBC=50^ BEC+ EBC+ ECB=180^90^+70^+ ECB=180^ ECB=20^ Then, we take triangle BFC, and find BFC=180^-50^-20^=110.