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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 $$\frac{12+\sqrt{15}}{5}$$.

##### Source of the problem

I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 2 from 2017

Plane Geometry

6 out of 10

##### Suggested Book

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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