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 $\angle BFC ?$

$90$

$100$

$105$

$110$

$120$

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.

Using the fact that the angles of a triangle add up to $180^{\circ},$ we get that $\begin{align*} \angle ECB &= 180^{\circ} - 70^{\circ} - 90^{\circ}\\&= 20^{\circ} \end{align*}$ and $\begin{align*} \angle DBC &= 180^{\circ} - 40^{\circ} - 90^{\circ}\\&= 50^{\circ}.\end{align*}$

Now, from $\triangle BFC,$ we get that $\begin{align*} \angle BFC &= 180^{\circ} - 50^{\circ} - 20^{\circ} \\&= 110^{\circ}.\end{align*}$ Thus, D is the correct answer.

2019 AMC 10A Problem 13

Solution:

Problem 13 in Other Years