# 1968 AHSME Problems/Problem 6

## Problem

Let side $AD$ of convex quadrilateral $ABCD$ be extended through $D$, and let side $BC$ be extended through $C$, to meet in point $E.$ Let $S$ be the degree-sum of angles $CDE$ and $DCE$, and let $S'$ represent the degree-sum of angles $BAD$ and $ABC.$ If $r=S/S'$, then:

$\text{(A) } r=1 \text{ sometimes, } r>1 \text{ sometimes}\quad\\ \text{(B) } r=1 \text{ sometimes, } r<1 \text{ sometimes}\quad\\ \text{(C) } 01\quad \text{(E) } r=1$

## Solution

$[asy] draw((-16,0)--(16,-24)); draw((0,24)--(16,-24)); draw((-16,0)--(0,24)); draw((0,-12)--(8,0)); dot((16,-24)); label("E",(16,-24),SE); dot((-16,0)); label("A",(-16,0),W); dot((0,24)); label("B",(0,24),N); dot((0,-12)); label("D",(0,-12),SW); dot((8,0)); label("C",(8,0),NE); [/asy]$

Because $ABCD$ is a convex quadrilateral, the sum of its interior angles is $360^{\circ}$. Thus, $S'=\measuredangle BAD + \measuredangle ABC=360^{\circ}-(\measuredangle BCD+\measuredangle ADC)$. Furthermore, because $\angle BCD$ and $\angle ADC$ are supplementary to $\angle DCE$ and $\angle CDE$, respectively, the four angles sum to $2*180^{\circ}=360^{\circ}$, so $\measuredangle BCD+\measuredangle ADC=360^{\circ}-(\measuredangle DCE+\measuredangle CDE)=360^{\circ}-S$. Plussing this expression for $\measuredangle BCD+\measuredangle ADC$ into the first equation, we see that $S'=360^{\circ}-(360^{\circ}-S)=S$, so $r=\frac{S}{S'}=1$, which is answer choice $\fbox{E}$.