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)

Capture-1

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.

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