# 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

##### Topic

Plane Geometry

##### Difficulty Level

6 out of 10

##### Suggested Book

‘An excursion in Mathematics’ published by Bhaskaracharya Pratishthana

# Start with hints

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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