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Problem 4: Let $ABC$ be an acute triangle and $D, E, F$ the feet of its altitudes from $A, B, C$, respectively. The line through $D$ parallel to $EF$ meets line $AC$ and line $AB$ at $Q$ and $R$, respectively. Let $P$ be the intersection of line $BC$ and line $EF$.
Prove that the circumcircle of $PQR$ passes through the midpoint of $BC$.

$M$ is the midpoint of side $BC$.

$\angle AQD=\angle AEF=\angle B$ and $\angle ARD=\angle AFE=\angle C$
\[\angle EPB=180\circ - \angle PEB-\angle EBP=180\circ-(\angle AEB-\angle AEF)-(\angle EBA+\angle ABP)\]
\[=180\circ-(90\circ-\angle B)-\{(90\circ-\angle A)+(180\circ-\angle B)\}=2\angle B+\angle A-180\circ=\angle B-\angle C\]

From $\Delta EBP$ \[\frac{PB}{sin\angle PEB}=\frac{BE}{sin\angle EPB}\Rightarrow \frac{PB}{sin(90\circ - \angle B)}=\frac{c\ sin \angle A}{sin (\angle B-\angle C)}\Rightarrow PB=\frac{c\ sin\angle A cos\angle B}{sin(\angle B-\angle C)}\]
$BD=c\ cos\angle B$
$\therefore PD=PB+BD=c\ cos\angle B\{\frac{sin(\angle B-\angle C)+sin\angle A}{sin(\angle B-\angle C)}\}$

From $\Delta ARD$ \[ \frac{RD}{sin\angle RAD}=\frac{AD}{sin\angle ARD}\Rightarrow \frac{RD}{sin(90\circ-\angle B)}=\frac{c\ sinB}{sin\angle C}\Rightarrow RD=\frac{c\ sin\angle B\ cos\angle B}{sin\angle C}\]
\[\therefore \frac{PD}{RD}=\frac{\{sin(\angle B-\angle C)+sin\angle A\}sin\angle C}{sin(\angle B-\angle C)sin \angle B}\]

From $\Delta ADQ$\[\ \frac{QD}{sin\angle DAQ}=\frac{AD}{sin\angle AQD}\Rightarrow \frac{QD}{sin(90\circ-\angle C)}=\frac{b\ sin\angle C}{sin\angle B}\]\[\Rightarrow QD=\frac{b\ cos\angle C sin\angle C}{sin\angle B}=\frac{a\ cos\angle C sin\angle C}{sin\angle A}\]
\[DM=BM-BD=\frac{a}{2}-c\ cos\angle B=\frac{a}{2}-\frac{a\ sin\angle C cos\angle B}{sin\angle A}=a\{\frac{sin\angle A-2sin\angle C cos\angle B}{2sin\angle A}\}\]
\[\therefore \frac{QD}{DM}=\frac{2sin\angle C cos\angle C}{sin\angle A-2sin\angle C cos\angle B}\]

Now,
\[\frac{PD}{RD}=\frac{QD}{DM}\]
\[\Leftrightarrow \frac{\{sin(\angle B-\angle C)+sin\angle A\}sin\angle C}{sin(\angle B-\angle C)sin \angle B}=\frac{2sin\angle C cos\angle C}{sin\angle A-2sin\angle C cos\angle B}\]
\[\Leftrightarrow \frac{sin(\angle B-\angle C)+sin\angle A}{sin(\angle B-\angle C)sin \angle B}=\frac{2cos\angle C}{sin\angle A-2sin\angle C cos\angle B}\]
\[\Leftrightarrow \frac{sin(\angle B-\angle C)+sin\angle A}{sin(\angle B-\angle C)sin \angle B}=\frac{2cos\angle C}{sin(\angle B+\angle C) -2sin\angle C cos\angle B}\]
\[\Leftrightarrow \frac{sin(\angle B-\angle C)+sin\angle A}{sin(\angle B-\angle C)sin \angle B}=\frac{2cos\angle C}{sin(\angle B-\angle C)}[\text {as} EF \text{intersects} BC, \angle B \ne \angle C\]
\[\Leftrightarrow \frac{sin(\angle B-\angle C)+sin\angle A}{sin \angle B}=2cos\angle C\]
\[\Leftrightarrow sin(\angle B-\angle C)+sin\angle (\angle B+\angle C)=2cos\angle C sin \angle B\]

Which is true.
$\therefore \frac{PD}{RD}=\frac{QD}{DM}\Rightarrow PD\cdot DM=QD\cdot RD$

So, $P,R,M,Q$ cyclic

for $P,Q,M,R$ to be cyclic we need to prove that $RD.DQ=PD.DM$.it can be easily deduced that $FD=RD=bcosB$
and $DE=DQ=ccosC$.we can also get $DM=\frac{1}{2}a-BD=bcosC-ccosB$.
so it suffices to prove that $PD=\frac{2bccosBcosC}{bcosC-ccosB}$.
applying sine law on $\triangle PDF$ yields us the desired figure.

why $\angle AEF=\angle B$ -tanmoy

tanmoy wrote:why $\angle AEF=\angle B$ -tanmoy

Note that the quadrilateral EFBC is cyclic, because <BEC = 90 = <BFC. Hence the equality.

We will have to prove that $PD.DC=RD.DQ$
$\angle{AFE}=\angle{BRQ}=\angle{BCQ}$
So, $BRCQ$ is cyclic.
Then, $RD.DQ=BD.DC$
We wanna prove, $PD.DM=BD.DC$
Or, $(PB+a/2-DM)DM=(a/2-DM)(a/2+DM)$
Or, $BP.DM=a^2/4-a/2.DM$
We get $(P,D;B,C)$ is a harmonic bundle.
So, $BP.CD=BD.CP$
Or, $BP(a/2+DM)=(a/2-DM)(BP+a)$
Or, $BP.DM=a^2/4-a/2.DM$
So, we are done...

Let M be the midpoint of BC. Now,M,D,E,F are cyclic (nine point circle). So, PD.PM=PF.PE=PB.PC=PM^2 -BM^2 (as BFEC is cyclic).
So,PM.PD=PM^2 - BM^2
PM^2- PD.DM - DM^2=PM^2-BM^2
=> PD.DM=BM^2 - DM^2=BD.CD

Now,<RCB= <RCD= <ACD= <EHA= <EFA= <RQA= <RQB. So, RCQB is cyclic. So, QD.RD=BD.CD=PD.DM
So, PQMR is cyclic.