# 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 $\triangle ABD.$ We can use $AB$ as the base and the altitude from $D$. Let's call the foot of the altitude $E.$ We have $\triangle BDE$ similar to $BAC.$

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