16) Let $A B C$ be an acute triangle and let $H$ on side $A B$ be the foot of the altitude from vertex $C$. Suppose that the area of triangle $A H C$ is to that of triangle $A B C$ as $\overline{A C}$ is to $2 \overline{A B}$. Which of the following statements is definitely true?
(A) $A B C$ is a right triangle (B) $\widehat{C A B}=60^{\circ}$
(D) $\overline{A B}=\overline{A C}$
(E) $A B C$ is equilateral
(C) $\overline{A B}=2 \overline{A H}$

![](https://cdn.mathpix.com/cropped/2024_04_17_62915d10d7246845095ag-2.jpg?height=227&width=343&top_left_y=998&top_left_x=1115)
16. The answer is $(\mathbf{B})$.

The area of $A H C$ is given by $(A H \cdot C H) / 2$ while the area of $A B C$ is given by $(A B \cdot C H) / 2$. Therefore, the ratio between the area of $A H C$ and that of $A B C$ is $\frac{A H}{A B}$; from the problem, we know that

$$
\frac{A H}{A B}=\frac{A C}{2 A B}
$$

and from this equality, we get $A C=2 A H$; this implies that the angle $\widehat{C A B}$ measures $60^{\circ}$. [Problem proposed by S. Mongodi]