Determine the length of **DC** in terms of $l_1$ and $l_2$

Question

In the given figure, E is the midpoint of the arc ABEC and ED is perpendicular to the chord BC at D. If the length of the chord AB is $l_1$, and that of BD is $l_2$, determine the length of DC in terms of $l_1$ and $l_2$

Source

Answer 1

The answer is $DC=l_1+l_2$.

Indeed, let $\theta=\angle EBC=\angle EAC$. We have also $\theta=\angle ECA$. Using the "sine law" we have $$\frac{AC}{\sin2\theta}=\frac{EC}{\sin\theta}$$ Thus $AC=(2\cos\theta) EC$. Also $\cos\theta=\dfrac{BD}{BE}$, so $l_2=(\cos\theta) BE$. This implies that $$AC\cdot BE=2l_2 EC\tag{1}$$ Now, since $ACEB$ is a cyclic quadrilateral, we have $$ BC\cdot AE=AC\cdot BE+ AB\cdot EC $$ Using (1) and recalling that $AE=EC$ we get $$ BC\cdot EC=2l_2\cdot EC+ l_1\cdot EC $$ Simplifying, we get $BC=2 l_2+l_1$, so $CD=l_2+l_1$. $\qquad\square$