2019 AMC 10A Problem 13

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 $\boxed{\textbf{(D)}\ 110}$.

A circle on a side as diameter is a signal for right angles: every other point of the circle sees that side at $90^\circ$, so the "other intersections" are the feet of altitudes. The intersection of two altitudes is the orthocenter $H$, and the general fact $\angle BHC = 180^\circ - \angle A$ (the two altitudes and the sides at $A$ form a quadrilateral with two right angles) gives $110^\circ$ at once.

• Misidentifying the diagonals of BCDE (they are BD and CE, not BE and CD) or using angle C instead of angle A in the orthocenter formula.
• Computing $\angle BFC$ as $180^\circ - 40^\circ = 140^\circ$ or as $90^\circ + 40^\circ / 2$ from a half-remembered incenter formula.