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

As shown in the diagram, $$ABCD$$ is a parallelogram where $$DC$$ is tangent to the circumcircle of $$\triangle ABC$$ which intersects $$AD$$ at $$E.$$ If $$AB = 4$$ and $$BE = 5,$$ find the length of $$DE.$$ Firstly, I noted that Power of Point could be used in this problem. I let $$DE = x$$ and I setup the equation $$x \cdot (x+AE) = 16,$$ since $$AB = CD.$$ However, from here, I got stuck as I do not know how to use the fact that $$BE = 5.$$ Can somebody help me?

• Try Angle Chasing Dec 19, 2020 at 4:38

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 $$DE=\frac{DC^2}{DA} = \frac{AB^2}{EB} = \frac{16}5$$