Question

Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}.$ Construct the circle with diameter $\overline{BC},$ and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB},$ respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE.$ What is the degree measure of $\angle BFC ?$

$90$

$100$

$105$

$110$

$120$

Answer: D

Concepts:inscribed angle angle chasing isosceles triangle

Difficulty rating: 1420

Solution:

Since $\overline{BC}$ is the diameter of the circle, we get that $\angle BDC$ and $\angle BEC$ are right angles.

We know that $\angle ABC = 70^{\circ}$ from the fact that $\triangle ABC$ is isosceles.

In $\triangle BCE$ and $\triangle BCD$ , respectively, $\begin{aligned} \angle ECB&=180^{\circ}-70^{\circ}-90^{\circ}\\ &=20^{\circ} \end{aligned}$ and $\begin{aligned} \angle DBC&=180^{\circ}-40^{\circ}-90^{\circ}\\ &=50^{\circ}. \end{aligned}$

Because $F$ lies on $BD$ and $CE$ , the other two angles of $\triangle BFC$ are $50^{\circ}$ and $20^{\circ}$ . Hence $\begin{aligned} \angle BFC&=180^{\circ}-50^{\circ}-20^{\circ}\\ &=110^{\circ}. \end{aligned}$ Thus, D is the correct answer.

2019 AMC 10A Problem 13

Answer: D

Difficulty rating: 1420

Solution:

Problem 13 in Other Years