$\triangle ABC$ is a triangle. $$|AE|=|AC|$$ $$AE\perp ED, EH\perp AC$$ $$|EC|=4|EB|$$ $$3|CH|=2|AH|$$ $$|EH|=22\;cm$$ $$|DE|=x\;cm$$ Find $x$.
My solution:
Let $|CH|=2k$. Then $|AH|=3k$, $|AE|=5k$.
By Pythagorean theorem on $\triangle AEH$, $|AE|=\frac{55}2$, $AH=\frac{33}{2}$ and hence $\cos\angle A=\frac 35$.
By law of cosines on $\triangle AEC$, $|EC|=\frac{55}{\sqrt5}$ and hence $|BE|=\frac{55}{4\sqrt5}$.
By half -angle formula $\cos\angle BED=\sin\angle AEC=\cos\frac{\angle A}2=\sqrt{\frac{\cos\angle A +1}{2}}=\frac2{\sqrt 5}.$
Edit: Here I made a mistake. I couldn't solve the question in time.:(
This is a test question sent to me from a student and time allocated for it is 90 seconds. I found my solution too slow. I wonder if somebody can find a faster way.
Thanks in advance.
Draw $CK\perp AE$
Since $\triangle ABC$ is isosceles $CK=EH=22$
Let $CK$ intersect $AB$ at $L$
then $DE\parallel CL$ ($\perp$ same line)
$\triangle ALK \sim \triangle ADE$ (AAA similarity)
$\frac{LK}{DE}=\frac{AK}{AE}= \frac {3}{5}=\frac{LK}{x}$ and $LK=\frac{3x}{5}$
$\triangle BDE \sim \triangle BLC$ (AAA similarity)
$\frac{BE}{BC}=\frac{DE}{LC}=\frac{1}{5}=\frac {x}{\frac{3x}{5}+22}$
$x=5$