Vtyjkyuk

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.

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

Since triangle $ABD$ is a right triangle, we have
$$AD=sqrtAB^2+BD^2=sqrt8^2+6^2=sqrt100=10$$
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
$$x=ADsintheta=10sintheta$$
where $theta=angle DAC$.

Since $BC=CD$, it follows that $angle DAB=2theta$, hence, since triangle $ABD$ is a right triangle, we get
beginalign*
&cos 2theta=fracABAD=frac45\[4pt]
implies;&1-2sin^2 theta=frac45\[4pt]
implies;&sin^2 theta=frac110\[4pt]
implies;&100sin^2 theta=10\[4pt]
implies;&(10sintheta)^2=10\[4pt]
implies;&x^2=10\[4pt]
endalign*
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 $AE,parallelBC$, quadrileral $EACB$ is a cyclic trapezoid, hence $EC=AB=8$, and $EB=AC=y$.

Then by Ptolemy's theorem, we get
beginalign*
&(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=frac3385\[4pt]
endalign*
Therefore $AE^2=largefrac3385$.

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, $angle HBC=arcsinleft(frac1sqrt10right)$ and $angle ABC=fracpi2+arcsinleft(frac1sqrt10right)$, and since $AC||AE$, then $angle BAE=pi-arcsinleft(frac1sqrt10right)$

We know that $angle BAD=arcsinleft(frac610right)$, thus $angle DAE=fracpi 2-arcsinleft(frac35right)-arcsinleft(frac1sqrt10right)$

Since we know that $triangle AED$ is right, therefore we know that $AE$ should be:
$$cos angle DAE=fracAEADimplies AE=ADcosangle DAE\
AE=10cosleft(fracpi 2-arcsinleft(frac35right)-arcsinleft(frac1sqrt10right)right)\$$
Using angle-sum identity for $sin$ and keeping in mind that $cos(arcsin(x))=sqrt1-x^2$, we get that the equation above evaluates to:
$$
AE=10cdotsin left(arcsinleft(frac35right)+arcsinleft(frac1sqrt10right)right)=13 sqrtfrac25$$
And thus:
$$bbox[10px, border:2px black solid]therefore AE^2=left(13 sqrtfrac25right)^2=frac3385$$

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 $sqrt5^2-3^2=4$ 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 $BC = CD = sqrt10$.

Let's also draw $BG$ perpendicular to $BC$ and $AE$ (and thus parallel to $DE$). Now, $AGB$ and $BFC$ are similar, so we get $AG = 1cdotfrac8sqrt10 = frac4sqrt105$ and $BG=3cdotfrac8sqrt10=frac12sqrt105$. 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 $BH=fracsqrt102$, so we have $OH=frac3sqrt102$. 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 $OI = BG - OH = frac12sqrt105 - frac3sqrt102 = frac9sqrt1010$. Now, finally, we can find $AI$ and thus $AE$ and thus $AE^2$: $AI = sqrt5^2-left(frac9sqrt1010right)^2=sqrt25-frac8110=sqrtfrac16910=frac13sqrt1010$, $AE=frac13sqrt105$, $AE^2=frac3385$