2019 AMC 10A Problem 13

1. Find the triangle's angles. Since $BC = AC$, the angles at $A$ and $B$ are equal. They share $180^\circ - 40^\circ = 140^\circ$, so $\angle A = \angle B = 70^\circ$.
2. Use the circle. A key fact: if $BC$ is a diameter of a circle and $X$ is any other point on that circle, then $\angle BXC = 90^\circ$. Both $D$ and $E$ are on this circle, so
$$\angle BDC = 90^\circ \quad\text{and}\quad \angle BEC = 90^\circ.$$
3. What this means. $D$ is on side $AC$ and $BD$ meets $AC$ at a right angle, so $BD$ is the height from $B$. In the same way, $CE$ is the height from $C$ to side $AB$.
4. Find the diagonals. Go around the quadrilateral in the order $B \to C \to D \to E$. The diagonals join opposite corners: $B$ with $D$, and $C$ with $E$. So $F$ is where $BD$ and $CE$ cross.
5. Angle $FBC$. In the right triangle $BDC$, the angle at $C$ is $40^\circ$, so the angle at $B$ is $90^\circ - 40^\circ = 50^\circ$.
6. Angle $FCB$. In the right triangle $BEC$, the angle at $B$ is $70^\circ$, so the angle at $C$ is $90^\circ - 70^\circ = 20^\circ$.
7. Finish in triangle $BFC$.
$$\angle BFC = 180^\circ - 50^\circ - 20^\circ = 110^\circ.$$

The answer is $\boxed{\textbf{(D)}\ 110}$.

A circle drawn on a side as its diameter is a signal for right angles: every other point of that circle sees the side at $90^\circ$. So the extra points $D$ and $E$ are the feet of two heights of the triangle. After that it is just the fact that angles in a triangle add to $180^\circ$, used three times.

Look at the quadrilateral $AEFD$. It has right angles at $E$ and $D$ (from the heights) and the angle $70^\circ$ at $A$. The four angles of a quadrilateral add to $360^\circ$, so $\angle EFD = 360^\circ - 90^\circ - 90^\circ - 70^\circ = 110^\circ$. The angle $\angle BFC$ is directly opposite $\angle EFD$ where the two diagonals cross, so it is also $110^\circ$.

Since $BC = AC$, the base angles at $A$ and $B$ are equal: $\angle A = \angle B = \frac{180^\circ - 40^\circ}{2} = 70^\circ$.

$D$ and $E$ lie on the circle with diameter $BC$, so $\angle BDC = 90^\circ$ and $\angle BEC = 90^\circ$. Thus $BD$ is the altitude from $B$ to $AC$, and $CE$ is the altitude from $C$ to $AB$.

The quadrilateral $BCDE$ has vertices in the order $B, C, D, E$, so its diagonals are $BD$ and $CE$, and $F$ is where the two altitudes meet.

In right triangle $BDC$, $\angle DBC = 90^\circ - 40^\circ = 50^\circ$. In right triangle $BEC$, $\angle ECB = 90^\circ - 70^\circ = 20^\circ$. In triangle $BFC$,
$$\angle BFC = 180^\circ - 50^\circ - 20^\circ = 110^\circ.$$

The answer is $\boxed{\textbf{(D)}\ 110}$.