Finding the length of $DE$ given $AB = 4$ and $BE = 5$

Since $AE$ and $BC$ are parallel, $ABCE$ is a symmetric trapezium, implying $CE=4$ and $AC=5$. By Ptolemy's theorem $$(x+AE)AE+4^2=5^2\implies (x+AE)AE=9$$ With the equation you already have, this produces the result $x=16/5$ and $AE=9/5$.


Note $\angle DAB = \angle DCB$, which means the arcs $EB$ and $CB$ are equal due to the shared arc $EC$. Then, $EB = CB= DA$ and, per the power of point $$DE=\frac{DC^2}{DA} = \frac{AB^2}{EB} = \frac{16}5$$

Tags:

Geometry

Power Of The Point

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