2019 AMC 10A Problem 13

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

Angles subtended by a diameter are right angles

Use the isosceles angles of

\triangle ABC

to chase the remaining angles

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.

Problem 13 in other years