# 2006 AMC 8 Problems/Problem 19

## Problem

Triangle $ABC$ is an isosceles triangle with $\overline{AB}=\overline{BC}$. Point $D$ is the midpoint of both $\overline{BC}$ and $\overline{AE}$, and $\overline{CE}$ is 11 units long. Triangle $ABD$ is congruent to triangle $ECD$. What is the length of $\overline{BD}$? $[asy] size(100); draw((0,0)--(2,4)--(4,0)--(6,4)--cycle--(4,0),linewidth(1)); label("A", (0,0), SW); label("B", (2,4), N); label("C", (4,0), SE); label("D", shift(0.2,0.1)*intersectionpoint((0,0)--(6,4),(2,4)--(4,0)), N); label("E", (6,4), NE);[/asy]$ $\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.5\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 5.5\qquad\textbf{(E)}\ 6$

## Solution

Since triangle $ABD$ is congruent to triangle $ECD$ and $\overline{CE} =11$, $\overline{AB}=11$. Since $\overline{AB}=\overline{BC}$, $\overline{BC}=11$. Because point $D$ is the midpoint of $\overline{BC}$, $\overline{BD}=\frac{\overline{BC}}{2}=\frac{11}{2}=\boxed{\textbf{(D)}\ 5.5}$.

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