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 $\overline{BC}$ , $\overline{BD}=\frac{\overline{BC}}{2}=\frac{11}{2}=\boxed{\textbf{(D)}\ 5.5}$ .

Video Solution

https://youtu.be/HzzzRJEppRA Soo, DRMS, NM

Video Solution by OmegaLearn

https://youtu.be/abSgjn4Qs34?t=2124

~ pi_is_3.14

Video Solution by WhyMath

https://youtu.be/bRJ_ghuH5qs

2006 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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