# Cyclic Quadrilaterals and Incenters (RMO 2015 Mumbai Region Solution)

Problem:  Let ABC be a right angled triangle with $$\angle B = 90^0$$ and let BD be the altitude from B on to AC. Draw $$DE \perp AB$$ and $$DF \perp BC$$. 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 $$\displaystyle { \Delta PFQ }$$ is similar to $$\displaystyle { \Delta RES }$$.

First note that $$\displaystyle { \Delta PFC \equiv \Delta BRE }$$ by simple angle chasing as follows:
$$\displaystyle { \angle PFC = \angle REB = 45^o}$$ (half of right angle as $$\displaystyle { DE \perp AB }$$ and RE bisects $$\displaystyle { \angle DEB}$$, etc. )
$$\displaystyle { \angle PCF = \angle RBE = \frac{\angle C} {2}}$$ as $$\displaystyle { \angle ABD = \angle C }$$ and BR is the angle bisector.

Since sides of similar triangles are proportional, hence $$\displaystyle { \frac {PF}{FC} = \frac {RE}{BE} }$$ — (i)

Similarly note that $$\displaystyle { \Delta QFB \equiv \Delta ASE }$$ by similar angle chasing.

Again as sides of similar triangles are proportional, hence $$\displaystyle { \frac {QF}{FB} = \frac {SE}{AE} }$$ — (ii)

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

$$\displaystyle { \frac{\frac {PF}{FC}}{\frac{QF}{FB}} = \frac{\frac {RE}{BE}}{\frac {SE}{AE}} }$$

$$\displaystyle { \Rightarrow \frac {PF}{FC} \times \frac{FB}{QF} = \frac {RE}{BE} \times \frac {AE}{SE} }$$

$$\displaystyle { \Rightarrow \frac {PF}{QF} \times \frac{FB}{FC} = \frac {RE}{SE} \times \frac {AE}{BE} }$$ — (iii)

Finally note that DE parallel to BC we have $$\displaystyle { \frac{AE}{BE} = \frac {AD}{DC} }$$

Also as DF parallel to BA we have $$\displaystyle { \frac{FB}{FC} = \frac {AD}{DC} }$$

Combining these two results we have $$\displaystyle { \frac{FB}{FC} = \frac {AE}{BE} }$$

Applying this to result (iii) we have
$$\displaystyle { \frac {PF}{QF} = \frac {RE}{SE} }$$
Also $$\displaystyle { \angle RES = \angle PFQ = 90^o }$$ as each is sum of $$\displaystyle { 45^o }$$

As pairs of sides of $$\displaystyle { \Delta PFQ, \Delta RES }$$ are proportional and included angle equal, hence the two triangles are similar hence equiangular.

Thus $$\displaystyle { \angle FPQ = \angle SRE = x }$$ (say).

We will use this result and little angle chasing to show

$$\displaystyle { \angle QPD + \angle QRD = 180^o }$$ leading to the conclusion that PQRD is cyclic.

As FD is parallel to BA $$\displaystyle { \angle FDC = \angle A }$$ (corresponding angles), thus $$\displaystyle { \angle PDC = \frac{\angle A}{2} }$$

Also $$\displaystyle { \angle PCD = \frac{\angle C}{2} }$$

Hence $$\displaystyle { \angle DPC = 180^o – \frac{\angle A}{2} = \frac{C}{2} = 135^o }$$ as $$\displaystyle { \frac{\angle A}{2} + \frac{\angle C}{2} = \frac{90^o}{2}}$$
Similarly $$\displaystyle { \angle CPF = 180^o – \frac{\angle C}{2} – 45^o = 135^o – \frac{\angle C}{2} }$$

$$\displaystyle { \angle QPD = 360^o – \angle CPF – \angle DPC -\angle FPQ = 360^o – (135^o – \frac{\angle C}{2}) – 135^o – x)= 90^o + \frac{\angle C}{2} – x }$$ (iv)

Similarly angle chasing gives $$\displaystyle { \angle QRD = 180^o – \angle DRS = 180^o – (180^o – \angle RDE -\angle RED – \angle RES) }$$

or $$\displaystyle { \angle QRD = \angle RDE + \angle RED + \angle RES = \frac{\angle A}{2} + 45^o + x }$$ –(v)

Now adding $$\displaystyle { \angle QRD + \angle QPD }$$ from (iv) and (v) we have

$$\displaystyle { \angle QRD + \angle QPD }$$
$$\displaystyle { = 90^o + \frac{\angle C}{2} – x +\frac{\angle A}{2} + 45^o + x}$$
$$\displaystyle { = 90^o +\frac{\angle C}{2}+\frac{\angle A}{2} + 45^o }$$
$$\displaystyle { = 90^o + 45^o + 45^o = 180^o}$$

Concluding PQRD is cyclic.