Skip to main content

2019 AMC 10A Problem 13

Problem 13 of 25IntermediateGeometry

Let △ABC\triangle ABC be an isosceles triangle with BC=ACBC = AC and ∠ACB=40∘.\angle ACB = 40^{\circ}. Construct the circle with diameter BC‾,\overline{BC}, and let DD and EE be the other intersection points of the circle with the sides AC‾\overline{AC} and AB‾,\overline{AB}, respectively. Let FF be the intersection of the diagonals of the quadrilateral BCDE.BCDE. What is the degree measure of ∠BFC?\angle BFC ?

Answer choices

Show solution

Solution

Since BC‾\overline{BC} is the diameter of the circle, we get that ∠BDC\angle BDC and ∠BEC\angle BEC are right angles. We know that ∠ABC=70∘\angle ABC = 70^{\circ} from the fact that △ABC\triangle ABC is isosceles. In △BCE\triangle BCE and △BCD\triangle BCD, respectively, ∠ECB=180∘−70∘−90∘=20∘ \begin{aligned} \angle ECB&=180^{\circ}-70^{\circ}-90^{\circ}\\ &=20^{\circ} \end{aligned} and ∠DBC=180∘−40∘−90∘=50∘. \begin{aligned} \angle DBC&=180^{\circ}-40^{\circ}-90^{\circ}\\ &=50^{\circ}. \end{aligned} Because FF lies on BDBD and CECE, the other two angles of △BFC\triangle BFC are 50∘50^{\circ} and 20∘20^{\circ}. Hence ∠BFC=180∘−50∘−20∘=110∘. \begin{aligned} \angle BFC&=180^{\circ}-50^{\circ}-20^{\circ}\\ &=110^{\circ}. \end{aligned} Thus, D is the correct answer.
AoPS wiki

Tagged: inscribed angle · angle chasing · isosceles triangle

More practice