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)

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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