Right angled triangle in a circle (B.Stat 2017, subjective 2)

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

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

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