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## Understand the problem

Consider a circle of radius 6 as given in the diagram below. Let $B,C,D$ and $E$ be points on the circle such that $BD$ and $CE$, when extended, intersect at $A$. If $AD$ and $AE$ have length 5 and 4 respectively, and $DBC$ is a right angle, then show that the length  of $BC$ is $\frac{12+\sqrt{15}}{5}$. Diagram : click here.
##### Source of the problem
I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 2 from 2017
Plane Geometry

##### Difficulty Level
6 out of 10

Do you really need a hint? Try it first!

Given, $AD=5, AE=4$ and $\angle DBC=90^\circ$. As $D,B,C$ are points on the circle having radius 6. Therefore $DC$ is the diameter of the circle $\Rightarrow DC=6×2=12$.

Now  DC is diameter of the circle $\Rightarrow \angle DEC=90^\circ$. Therefore $\angle DEA$ is also right angle.

The length of $DE=\sqrt{5^2-4^2}=3$.  And the length of $EC=\sqrt{12^2-3^2}=3\sqrt{15}$. Therefore $AC=AE+EC=4+3\sqrt{15}$.

From $∆AED$ and $∆ABC$ we have, $\frac{AD}{DE}=\frac{AC}{BC} \Rightarrow BC=\frac{AC\cdot DE}{AD}=\frac{(4+3\sqrt{15})\cdot 3}{5}$. Therefore the length of $BC$ is $\frac{12+9\sqrt{15}}{5}$.   (Ans.)

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