# 2017 AMC 8 Problems/Problem 16

## Problem 16

In the figure below, choose point $D$ on $\overline{BC}$ so that $\triangle ACD$ and $\triangle ABD$ have equal perimeters. What is the area of $\triangle ABD$? $[asy]draw((0,0)--(4,0)--(0,3)--(0,0)); label("A", (0,0), SW); label("B", (4,0), ESE); label("C", (0, 3), N); label("3", (0, 1.5), W); label("4", (2, 0), S); label("5", (2, 1.5), NE);[/asy]$

$\textbf{(A) }\frac{3}{4}\qquad\textbf{(B) }\frac{3}{2}\qquad\textbf{(C) }2\qquad\textbf{(D) }\frac{12}{5}\qquad\textbf{(E) }\frac{5}{2}$

## Solution

We know that the perimeters of the two small triangles are $3+CD+AD$ and $4+BD+AD$. Setting both equal and using $BD+CD = 5$, we have $BD = 2$ and $CD = 3$. Now, we simply have to find the area of $\triangle ABD$. Since $\frac{BD}{CD} = \frac{2}{3}$, we must have $\frac{[ABD]}{ACD]} = 2/3. Combining this with the fact that$[ABC] = [ABD] + [ACD] = \frac{3*4}{2} = 6$, we get$[ABD] = \frac{2}{5}[ABC] = \frac{2}{5} * 6 = \boxed{\textbf{(D) }$\frac{12}{5}}$