# Difference between revisions of "1967 AHSME Problems/Problem 32"

## Problem

In quadrilateral $ABCD$ with diagonals $AC$ and $BD$, intersecting at $O$, $BO=4$, $OD = 6$, $AO=8$, $OC=3$, and $AB=6$. The length of $AD$ is: $\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$

## Solution

We note that $BO \cdot DO = AO \cdot CO = 24$. This is the Power of a Point Theorem which only happens to chords in circles. Therefore, we conclude that $ABCD$ is cyclic. We can proceed with similar triangles. Because of inscribed angles, $\triangle ABO \simeq \triangle DCO$ and $\triangle ADO \simeq \triangle BCO$. We find $\frac{CD}{AB} = \frac{3}{4} \implies CD = \frac{9}{2}$ with the first similarity and $\frac{BC}{AD} = \frac{3}{6} \implies BC = \frac{AD}{2}$ with the second similarity. Now, we can apply Ptolemy's theorem which states that in a cyclic quadrilateral, $AB \cdot CD + AD \cdot BC = AC \cdot BD$. We can plug in out values to get $6 \cdot \frac{9}{2} + AD \cdot \frac{AD}{2} = 11 \cdot 10 = 110$. We solve for $AD$ to get $AD = \boxed{\textbf{(E) } \sqrt{166}}$. - [b] lucasxia01[/b]

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