# Difference between revisions of "1995 AJHSME Problems/Problem 13"

## Problem

In the figure, $\angle A$, $\angle B$, and $\angle C$ are right angles. If $\angle AEB = 40^\circ$ and $\angle BED = \angle BDE$, then $\angle CDE =$ $[asy] dot((0,0)); label("E",(0,0),SW); dot(dir(85)); label("A",dir(85),NW); dot((4,0)); label("D",(4,0),SE); dot((4.05677,0.648898)); label("C",(4.05677,0.648898),NE); draw((0,0)--dir(85)--(4.05677,0.648898)--(4,0)--cycle); dot((2,2)); label("B",(2,2),N); draw((0,0)--(2,2)--(4,0)); pair [] x = intersectionpoints((0,0)--(2,2)--(4,0),dir(85)--(4.05677,0.648898)); dot(x); dot(x); label("F",x,SE); label("G",x,SW); [/asy]$ $\text{(A)}\ 75^\circ \qquad \text{(B)}\ 80^\circ \qquad \text{(C)}\ 85^\circ \qquad \text{(D)}\ 90^\circ \qquad \text{(E)}\ 95^\circ$

## Solution

Because $\angle BED=\angle BDE$, $\angle B=90^\circ$, and $\triangle BED$ is a triangle, we get: $$\angle B + \angle BED +\angle BDE=180$$ $$90 +\angle BED +\angle BED=180$$ $$2\angle BED=90$$ $$\angle BED=\angle BDE=45^\circ$$

So $\angle AED=\angle AEB +\angle BED=40 +45=85^\circ$ Since ACDE is a quadrilateral, the sum of its angles is 360. Therefore: $$\angle A +\angle C +\angle CDE +\angle AED=360$$ $$90 +90 +\angle CDE +85=360$$ $$\angle CDE=95^\circ \text{(E)}$$

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