This is a problem from RMO 2015 from Mumbai Region based on Cyclic Quadrilaterals and Incenters.

**Problem: RMO 2015 Mumbai Region**

Let ABC be a right angled triangle with and let BD be the altitude from B on to AC. Draw and . Let P, Q, R and S be respectively the incenters of triangle DFC, DBF, DEB and DAE. Suppose S, R, Q are collinear. Prove that P, Q, R, D lie on a circle.

**Discussion: (Diagram courtesy Eeshan Banerjee)**

We will show is similar to .

First note that by simple angle chasing as follows:

(half of right angle as and RE bisects , etc. )

as and BR is the angle bisector.

Since sides of similar triangles are proportional, hence — (i)

Similarly note that by similar angle chasing.

Again as sides of similar triangles are proportional, hence — (ii)

Therefore combining (i) and (ii) we have:

— (iii)

Finally note that DE parallel to BC we have

Also as DF parallel to BA we have

Combining these two results we have

Applying this to result (iii) we have

Also as each is sum of

As pairs of sides of are proportional and included angle equal, hence the two triangles are similar hence equiangular.

Thus (say).

We will use this result and little angle chasing to show

leading to the conclusion that PQRD is cyclic.

As FD is parallel to BA (corresponding angles), thus

Also

Hence as

Similarly

(iv)

Similarly angle chasing gives

or –(v)

Now adding from (iv) and (v) we have

Concluding PQRD is cyclic.

## Chatuspathi:

**Paper:**RMO 2015 (Mumbai Region)**What is this topic:**Geometry**What are some of the associated concepts:**Similarity of triangles, cyclic quadrilateral, angle chasing**Where can learn these topics:**Cheenta**Book Suggestions:**Challenges and Thrills of Pre-College Mathematics by Venkatchala

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