Try this beautiful problem from ISI BStat 2017 Subjective 2 based on right-angled triangle in a circle.

Problem: ISI BStat 2017 Subjective 2

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 $$ \frac {12 + 9 \sqrt {15} }{5} $$

I.S.I. 2017 geometry problem

Discussion:

This is a simple application of the power of a point. We have to find the power of point A.

Suppose B = (0,0), C = (x,0) and D = (0,y).

Since AD = 5, therefore, coordinate of A = (0, 5+y).

Using the property of the power of point A, we have: ( AD \times AB = AE \times AC ) . Since AE is 4, this implies

DEC is right angled at E so is DEA. In triangle DEA using Pythagoras DE=3. NOW ADE & ABC ARE SIMILAR SO , ARE AD/DE=AC/BC. AND AC=5+ROOT(12^2+5^2)=>4+3ROOT15 USING THE RATIO WE GET 4+3ROOT15/BC=5/3 BC=(12+9ROOT15)/5

There is a more elegant solution available. Using secant theorem sinc=4/5

Hence cosc=3/5.

Also dec is right angled at E so is DEA. In triangle DEA using pythagoras DE=3

In triangle DEC, using pythagoras EC= 3root15.

Hence BC=AC cosc= (4+3root15)*3/5

Proved

sir i have a solution just using classic euclidean geometry.how to send ?

DEC is right angled at E so is DEA. In triangle DEA using Pythagoras DE=3.

NOW ADE & ABC ARE SIMILAR SO , ARE AD/DE=AC/BC. AND AC=5+ROOT(12^2+5^2)=>4+3ROOT15

USING THE RATIO WE GET 4+3ROOT15/BC=5/3 BC=(12+9ROOT15)/5