# Cyclic Pentagon sides

In cyclic pentagon ABCDE, ∠ABD = 90◦ , BC = CD, and AE is parallel to BC. If AB = 8 and BD = 6, find $AE^2$ .

I can't find anything, I have been angle chasing the problem, but I can't find sufficient information to use cosine law. I can't find the length BE or the angle EBD in terms of the angles BAD and BDA.

• can you made an Image of this problem please? – Dr. Sonnhard Graubner Aug 13 '18 at 11:18
• Sorry, I dont think I can make an accurate image for this problem . I might confuse , there was no image given in the questionnaire, – SuperMage1 Aug 13 '18 at 11:20
• Since $\angle ABD=90$, then $AD$ is the diameter of the circle. By Pythagoras $AD=10$. Let $O$ be the center of the circle and midpoint of $AD$. Then $OC$ divides $BD$ in half. Apply Pythagoras twice to get that $BC=\sqrt{10}$. Now look at the trapezoid $ABCE$, you can compute its base $AE$ by Pythagoras. – user583185 Aug 13 '18 at 11:53
• $OC=5$ for being a radius. Suppose $OC$ intersects $BD$ at $F$. Then $OB=5$ for being a radius, and $FB=6/2=3$. Then $OF=4$ by Pythagoras. Therefore $FC=5-4=1$. By Pythagoras in the triangle $CFB$ you get $BC=\sqrt{FB^2+FC^2}=\sqrt{3^2+1^2}=\sqrt{10}$. – user583185 Aug 13 '18 at 12:11
• Two use Pythagoras on $ABCE$ draw the perpendicular $CG$ to $AE$. Then you have one Pythagoras on the triangle $CGE$. Draw the perpendicular $OH$ to $AE$. You get another Pythagoras on $OHE$. You can put as unknowns $AE$ and say the $OH$ to get a system of two equations and two unknowns. – user583185 Aug 13 '18 at 12:14

Since $\angle ABD=90^{\circ}$, it follows that $AD$ is a diameter.

Since triangle $ABD$ is a right triangle, we have $$AD=\sqrt{AB^2+BD^2}=\sqrt{8^2+6^2}=\sqrt{100}=10$$ Let $x=CD,\;y=AC,\;z=AE$.

Our goal is to find $z^2$.

First, we get $x^2$ . . .

Since $AD$ is a diameter, it follows that triangle $ACD$ is a right triangle, hence $$x=AD\sin\theta=10\sin\theta$$ where $\theta=\angle DAC$.

Since $BC=CD$, it follows that $\angle DAB=2\theta$, hence, since triangle $ABD$ is a right triangle, we get \begin{align*} &\cos 2\theta=\frac{AB}{AD}=\frac{4}{5}\\[4pt] \implies\;&1-2\sin^2 \theta=\frac{4}{5}\\[4pt] \implies\;&\sin^2 \theta=\frac{1}{10}\\[4pt] \implies\;&100\sin^2 \theta=10\\[4pt] \implies\;&(10\sin\theta)^2=10\\[4pt] \implies\;&x^2=10\\[4pt] \end{align*} Next, we get $y^2$ . . .

Since $AD$ is a diameter, triangle $ACD$ is a right triangle, hence $$y^2=AD^2-CD^2=10^2-x^2=100-10=90$$ Finally, we get $z^2$ . . .

Since $AE{\,\parallel}BC$, quadrileral $EACB$ is a cyclic trapezoid, hence $EC=AB=8$, and $EB=AC=y$.

Then by Ptolemy's theorem, we get \begin{align*} &(AE)(BC)+(AB)(EC)=(AC)(EB)\\[4pt] \implies\;&zx+8^2=y^2\\[4pt] \implies\;&zx+64=90\\[4pt] \implies\;&zx=26\\[4pt] \implies\;&z^2x^2=26^2\\[4pt] \implies\;&10z^2=676\\[4pt] \implies\;&z^2=\frac{338}{5}\\[4pt] \end{align*} Therefore $AE^2={\large{\frac{338}{5}}}$.

Since $\angle ABD$ is right, this means $\triangle ABD$ is an inscribed right triangle with $AD$ as the diameter of the circle $O$. This also means that $\angle AED$ is right since it is also an inscribed right triangle.

To satisfy $BC=CD$, $OC$ must be a perpendicular bisector of $BD$. Using Pythagorean theorem, we can workout the lengths of the segments:

With $AD=10$, then $OD=5$. With $HD=3$, we know that $HO=4$, thus $HC=1$, because $OC=OH+HC=5$.

Therefore, $\angle HBC=\arcsin\left(\frac{1}{\sqrt10}\right)$ and $\angle ABC=\frac\pi2+\arcsin\left(\frac{1}{\sqrt10}\right)$, and since $AC||AE$, then $\angle BAE=\pi-\arcsin\left(\frac{1}{\sqrt10}\right)$

We know that $\angle BAD=\arcsin\left(\frac{6}{10}\right)$, thus $\angle DAE=\frac{\pi }{2}-\arcsin\left(\frac{3}{5}\right)-\arcsin\left(\frac{1}{\sqrt{10}}\right)$

Since we know that $\triangle AED$ is right, therefore we know that $AE$ should be: $$\cos \angle DAE=\frac{AE}{AD}\implies AE=AD\cos\angle DAE\\ AE=10\cos\left(\frac{\pi }{2}-\arcsin\left(\frac{3}{5}\right)-\arcsin\left(\frac{1}{\sqrt{10}}\right)\right)\\$$ Using angle-sum identity for $\sin$ and keeping in mind that $\cos(\arcsin(x))=\sqrt{1-x^2}$, we get that the equation above evaluates to: $$AE=10\cdot\sin \left(\arcsin\left(\frac{3}{5}\right)+\arcsin\left(\frac{1}{\sqrt{10}}\right)\right)=13 \sqrt{\frac{2}{5}}$$ And thus: $$\bbox[10px, border:2px black solid]{\therefore AE^2=\left(13 \sqrt{\frac{2}{5}}\right)^2=\frac{338}5}$$

• $13^2×2\ne 388$, was $338$ meant? – Oscar Lanzi Aug 15 '18 at 0:58
• Yes, thank you @Oscar Lanzi – John Glenn Aug 15 '18 at 1:00

Since $\angle ABD$ is a right angle, we know that $AD$ is a diameter of the circle; by Pythagoras we know that the diameter is $10$ and thus the radius is $5$. since $E$ is also on the circle, $AED$ is a right triangle as well.

Chord $BD$ is $6$ units long and thus $\sqrt{5^2-3^2}=4$ units from the center, which we'll call $O$; since $BC = CD$ we know that $C$ is on the perpendicular bisector of $BD$, and since it's also on the circle, we know it's $5$ units from $O$. We'll place $F$ at the midpoint of $BD$; $BF = DF = 3$, $CF = 1$, and $BC = CD = \sqrt{10}$.

Let's also draw $BG$ perpendicular to $BC$ and $AE$ (and thus parallel to $DE$). Now, $AGB$ and $BFC$ are similar, so we get $AG = 1\cdot\frac{8}{\sqrt{10}} = \frac{4\sqrt{10}}{5}$ and $BG=3\cdot\frac{8}{\sqrt{10}}=\frac{12\sqrt{10}}{5}$. This last one is important!

Since $BC$ is a chord, its perpendicular bisector passes through $O$. We'll call the midpoint of $BC$ $H$, and we find that $BHO$ is a right triangle with $BO = 5$ and $BH=\frac{\sqrt{10}}{2}$, so we have $OH=\frac{3\sqrt{10}}{2}$. But $OH$ is parallel to $BG$! Let's extend $OH$ to meet $AE$ at $I$. since $BGIH$ is a parallelogram, $BG = HI$, so we can find $OI$, which tells us the distance of the chord $AE$ from the center. It's $OI = BG - OH = \frac{12\sqrt{10}}{5} - \frac{3\sqrt{10}}{2} = \frac{9\sqrt{10}}{10}$. Now, finally, we can find $AI$ and thus $AE$ and thus $AE^2$: $AI = \sqrt{5^2-\left(\frac{9\sqrt{10}}{10}\right)^2}=\sqrt{25-\frac{81}{10}}=\sqrt{\frac{169}{10}}=\frac{13\sqrt{10}}{10}$, $AE=\frac{13\sqrt{10}}{5}$, ${AE^2=\frac{338}{5}}$