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Solved Challenge Cyclic Quadrilateral

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,632
$ABCD$ is a cyclic quadrilateral such that $AB=BC=CA$. Diagonals $AC$ and $BD$ intersect at $E$. Given that $BE=19$ and $ED=6$, find all the possible values of $AD$.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,694
\begin{tikzpicture}

\draw circle (4) ;
\coordinate [label=left:$A$] (A) at (210:4) ;
\coordinate [label=above:$B$] (B) at (90:4) ;
\coordinate [label=right:$C$] (C) at (330:4) ;
\coordinate [label=below:$D$] (D) at (290:4) ;
\coordinate [label=above right:$E$] (E) at (intersection of A--C and B--D) ;
\draw (A) -- (B) -- node[ right ]{$x$} (C) -- (D) -- (A) --(C) ;
\draw (B) -- node[ right ]{$19$} (E) -- node[ right ]{$6$} (D) ;
\draw (-0.1,3.3) node{$\theta$} ;
\end{tikzpicture}
Let $x$ be the side length of the equilateral triangle $ABC$, and $\theta$ the angle $ABD$, as in the diagram.

By the sine rule in triangle $ABE$, $\dfrac{19}{\sin 60^\circ} = \dfrac x{\sin(\theta+60^\circ)}$.

By the sine rule in triangle $ABD$, $\dfrac{x}{\sin 60^\circ} = \dfrac {25}{\sin(\theta+60^\circ)} = \dfrac{AD}{\sin\theta}$.

Therefore $\dfrac{19}x = \dfrac x{25}$ and hence $x = 5\sqrt{19}$. Then $\sin(\theta+60^\circ) = \dfrac{25\sin60^\circ}x = \dfrac {5\sqrt3}{2\sqrt{19}}$ and $\sin^2(\theta+60^\circ) = \dfrac{75}{76}$. So $\cos^2(\theta+60^\circ) = \dfrac{1}{76}$ and $\cos(\theta+60^\circ) = \pm\dfrac1{2\sqrt{19}}.$ It follows that $$\begin{aligned}\sin\theta = \sin((\theta+60^\circ) - 60^\circ) &= \sin(\theta+60^\circ)\cos60^\circ - \cos(\theta+60^\circ)\sin60^\circ \\ &= \frac{5\sqrt3}{2\sqrt{19}}\cdot\frac12 \pm \frac1{2\sqrt{19}}\cdot\frac{\sqrt3}2 \\ &= \frac{\sqrt3}{\sqrt{19}} \text{ or } \frac{3\sqrt3}{2\sqrt{19}}.\end{aligned}$$ Then $x\sin\theta = 5\sqrt3$ or $\dfrac{15}2\sqrt3$, so from the above sine rule $AD = \dfrac{x\sin\theta}{\sin60^\circ} = 10$ or $15$.

The above diagram shows the longer alternative $AD = 15$, with $CD = 10$. The other alternative comes from interchanging $A$ and $C$.