# 1984 AHSME Problems/Problem 26

## Problem

In the obtuse triangle $ABC$ with $\angle C>90^\circ$, $AM=MB$, $MD\perp BC$, and $EC\perp BC$ ($D$ is on $BC$, $E$ is on $AB$, and $M$ is on $EB$). If the area of $\triangle ABC$ is $24$, then the area of $\triangle BED$ is

$\mathrm{(A) \ }9 \qquad \mathrm{(B) \ }12 \qquad \mathrm{(C) \ } 15 \qquad \mathrm{(D) \ }18 \qquad \mathrm{(E) \ } \text{Not uniquely determined}$

## Solution 1

We let side $BC$ have length $a$, $AB$ have length $c$, and $\angle ABC$ have angle measure $\beta$. We then have that

$$[ABC]=24=\frac{AB\cdot BC\sin{\angle ABC}}{2}=\frac{ac\sin{\beta}}{2}$$

Now I shall find the lengths of $BD$ and $CE$ in terms of the defined variables. Note that $M$ is defined to be the midpoint of $AB$, so $BM=\frac{c}{2}$. We can then use trigonometric manipulation on triangle $BDM$ to get that $BD=\frac{c\cos{\beta}}{2}$. We can also use trig manipulation on $BCE$ to get that $CE=a\tan{\beta}$.

Now note that $CE$ is the height of triangle $BDE$ originating from vertex $E$, so we have that

$[BED]=\frac{BD\cdot CE}{2}=\frac{ac\cos{\beta}\tan{\beta}}{4}=\frac{ac\sin{\beta}}{4}$

However, this is simply half the area of triangle $ABC$, so $[BED]=12$, which makes $\boxed{\textbf{B}}$ the correct answer.

## Solution 2

It is well known that $[CMB]=12$. But $\triangle{CBE}$~$\triangle{DBM}$. As a result $[CMB]=[BED] \implies \boxed{\textbf{B}}$.