2. For a triangle with vertices $A, B, C$, we know that $A B=5, B C=4$ and $A C=A M$, where $M$ is the midpoint of side $B C$. What is the length of side $A C$?
(A) 4
(B) $\sqrt{17}$
(C) $3 \sqrt{2}$
(D) $2 \sqrt{5}$
(E) $\sqrt{21}$
2. The answer is $\mathbf{( B )}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_04503179b337e2a9d877g-07.jpg?height=497&width=505&top_left_y=674&top_left_x=821)

Since $M$ is the midpoint, we get that $B M=M C=2$. Let $H$ be the midpoint of $M C$, and in particular $M H=H C=1$; since the triangle $A M C$ is isosceles with base $M C$, $H$ is also the foot of the altitude, and therefore the angle $\angle A H B$ is right.

Thus, the triangle $A B H$ is right-angled, with hypotenuse $A B$ of length 5 and leg $B H$ of length 3; it follows that $A H=4$. Finally, $A C$ is the hypotenuse of the triangle $A H C$, so $A C=\sqrt{4^{2}+1^{2}}=\sqrt{17}$.