# Difference between revisions of "1996 AHSME Problems/Problem 21"

## Problem

Triangles $ABC$ and $ABD$ are isosceles with $AB=AC=BD$, and $BD$ intersects $AC$ at $E$. If $BD$ is perpendicular to $AC$, then $\angle C+\angle D$ is $[asy] size(120); pair B=origin, A=1*dir(70), M=foot(A, B, (3,0)), C=reflect(A, M)*B, E=foot(B, A, C), D=1*dir(20); dot(A^^B^^C^^D^^E); draw(A--D--B--A--C--B); markscalefactor=0.005; draw(rightanglemark(A, E, B)); dot(A^^B^^C^^D^^E); pair point=midpoint(A--M); label("A", A, dir(point--A)); label("B", B, dir(point--B)); label("C", C, dir(point--C)); label("D", D, dir(point--D)); label("E", E, dir(point--E)); [/asy]$ $\text{(A)}\ 115^\circ\qquad\text{(B)}\ 120^\circ\qquad\text{(C)}\ 130^\circ\qquad\text{(D)}\ 135^\circ\qquad\text{(E)}\ \text{not uniquely determined}$

## Solution

Redraw the figure as a concave pentagon $ADECB$: $[asy] size(120); pair B=origin, A=1*dir(70), M=foot(A, B, (3,0)), C=reflect(A, M)*B, E=foot(B, A, C), D=1*dir(20); dot(A^^B^^C^^D^^E); draw(A--D--E--C--B--A); markscalefactor=0.005; draw(rightanglemark(D, E, C)); dot(A^^B^^C^^D^^E); pair point=midpoint(A--M); label("A", A, dir(point--A)); label("B", B, dir(point--B)); label("C", C, dir(point--C)); label("D", D, dir(point--D)); label("E", E, dir(point--E)); [/asy]$

The angles of the pentagon will still sum to $180^\circ \cdot 3 = 540^\circ$, regardless of whether the pentagon is concave or not. As a quick proof, note that the nine angles of three original triangles $\triangle AEB$, $\triangle CBE$, and $\triangle DEA$ all make up the angles of the pentagon without overlap.

Since reflex $\angle E = 270^\circ$, we have: $\angle C + \angle B + \angle A + \angle D = 540^\circ - 270^\circ = 270^\circ$.

From isosceles $\triangle ABC$, we get $\angle B = \angle C$, so: $2\angle C + \angle D + \angle A = 270^\circ$

From isosceles $/triangle ABD$, we get $\angle A = \angle D$, so: $2\angle C + 2\angle D = 270^\circ$

\$\angle C + \angle D = 135^\circ