# 1992 AHSME Problems/Problem 25

## Problem

In $\triangle{ABC}$, $\angle ABC=120^\circ,AB=3$ and $BC=4$. If perpendiculars constructed to $\overline{AB}$ at $A$ and to $\overline{BC}$ at $C$ meet at $D$, then $CD=$

$\text{(A) } 3\quad \text{(B) } \frac{8}{\sqrt{3}}\quad \text{(C) } 5\quad \text{(D) } \frac{11}{2}\quad \text{(E) } \frac{10}{\sqrt{3}}$

## Solution 1 (Extending Line Segments)

We begin by drawing a diagram. $[asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = A+dir(55); pair D = A+dir(0); pair B = extension(A,A+dir(90),C,C+dir(-155)); label("A",A,S); label("C",C,NE); label("D",D,SE); label("B",B,NW); label("4",B--C,NW); label("3",A--B,W); draw(A--C--D--cycle); draw(A--B--C); draw(rightanglemark(B,C,D,2)); draw(rightanglemark(B,A,D,2)); [/asy]$ We extend $CB$ and $DA$ to meet at $E.$ This gives us a couple right triangles in $CED$ and $BEA.$ $[asy] import olympiad; import cse5; import geometry; size(250); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = A+dir(55); pair D = A+dir(0); pair B = extension(A,A+dir(90),C,C+dir(-155)); pair E = extension(A,A+2*dir(180),B,B+2*dir(-155)); label("A",A,S); label("C",C,NE); label("D",D,SE); label("B",B,NW); label("4",B--C,NW); label("3",A--B,W); label("E",E,SW); draw(A--C--D--cycle); draw(A--B--C); draw(rightanglemark(B,C,D,2)); draw(rightanglemark(B,A,D,2)); draw(A--E--B,dashed); [/asy]$ We see that $\angle E = 30^\circ$. Hence, $\triangle BEA$ and $\triangle DEC$ are 30-60-90 triangles.

Using the side ratios of 30-60-90 triangles, we have $BE=2BA=6$. This tells us that $CE=BC+BE=4+6=10$. Also, $EA=3\sqrt{3}$.

Because $\triangle DEC\sim\triangle BEA$, we have $$\frac{10}{3\sqrt{3}}=\frac{CD}{3}.$$ Solving the equation, we have \begin{align*} \frac{CD}3&=\frac{10}{3\sqrt{3}}\\ CD&=3\cdot\frac{10}{3\sqrt{3}}\\ CD&=\frac{10}{\sqrt{3}}\ \end{align*} Hence, $CD=\boxed{\textbf{E}}$.

## Solution 2 (Cyclic Quadrilaterals, Right Triangles, LoC)

Since $\angle{A}+\angle{C} = 180^{\circ}, ABCD$ is cyclic. Using Ptolemy's Theorem gets $\text{(1): } 4AD + 3CD = \sqrt{37}BD.$

Right triangles $\Delta ABD$ and $\Delta CBD$ obtain $\text{(2): } 9+AD^2=BD^2$ and $\text{(3):} 16+CD^2=BD^2,$ respectively.

Seeing squares in $\text{(2)}$ and $\text{(3)}$, we square $\text{(1)}$ and get $\text{(4): } 16AD^2+9CD^2 + 24AD\cdot CD = 37BD^2.$

We don't like that $AD\cdot CD$ term, but fortunately LoC exists: $37 = AD^2 + CD^2 - 2*AD*CD*\cos(60^{\circ})$. Solving for $AD\cdot CD$ and plugging it into $\text{(4)}$, and using $AD^2 = 7 + CD^2$ and $BD^2 = 16 + CD^2$ from the first two equations, gets $40(CD^2+7)+33CD^2 - 24\cdot 37 = 37(16+CD^2).$

Solve for $CD = \boxed{\textbf{E}}$.

~PureSwag