Isn't it relatively simple though, for Advanced P4 at least, if the solution below works?

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Let $K$ and $L$ be the intersection of $AP$ with $(CDP)$ and $(BDP)$ and let $I$ be the incenter of $ABC$. Note that $BP^2=BD\cdot BC$ thus $BP$ is tangent to $(CKDP)$ and similarly $CP$ is tangent to $(BDLP)$. $$\measuredangle PDK=180-\measuredangle PCK= 90+\measuredangle CPK=180-\measuredangle BPK=\measuredangle BDL=90+\measuredangle BDM=180-\measuredangle PDM$$So $M,D,K$ are collinear and similarly $N,L,D$ are collinear. $$\measuredangle AMI=90-\measuredangle BDM=\measuredangle PCK= \measuredangle BPL$$thus $APMB$ is cyclic and similarly $APNC$ is cyclic. So $IM\cdot IB=IP\cdot IA=IN\cdot IC$. Hence $BMNC$ is cyclic and we are done $\square$ [asy][asy] import geometry; size(8cm); pair A,B,C,P,D,M,N,K,L,I; A=dir(130); B=dir(210); C=dir(330);P=intersectionpoints(circle(B,C),line(A,incenter(A,B,C)))[1];D=foot(P,B,C);M=incenter(A,B,D);N=incenter(A,C,D);K=foot(C,A,P);L=foot(B,A,P); I=incenter(A,B,C); draw(A--B--C--cycle);filldraw((A--P--M--B--cycle),palered);filldraw((A--P--N--C--cycle),paleblue); draw(circle(P,D,C),royalblue);draw(circle(P,D,B),red);draw(A--K);draw(C--K); draw(B--L);draw(P--D);draw(M--I--N);draw(D--N);draw(M--K); dot(P);dot(B);dot(C);dot(A);dot(D);dot(M);dot(N);dot(K);dot(L);dot(I); label("$A$",A,dir(A));label("$B$",B,dir(B));label("$C$",C,dir(C));label("$D$",D,dir(240));label("$K$",K,dir(B));label("$P$",P,dir(80)); label("$M$",M,dir(130));label("$N$",N,dir(60));label("$L$",L,dir(160));label("$I$",I,dir(140)); [/asy][/asy]

Here's another interesting fact about this problem. Let $E$ be the tangency point of the incircle of triangle $ABC$ with $BC$, suppose that there is a point $Q$ on the segment $AI$ such that $EQ$ has the same length as the tangent drawn from $A$ to the incircle of triangle $ABC$. Prove that $AD$ and $EQ$ are parallel.

This was one of my favorite geometry problems . Here is a sketch of my solution to the extension: XX-math-XX wrote: Here's another interesting fact about this problem. Let $E$ be the tangency point of the incircle of triangle $ABC$ with $BC$, suppose that there is a point $Q$ on the segment $AI$ such that $EQ$ has the same length as the tangent drawn from $A$ to the incircle of triangle $ABC$. Prove that $AD$ and $EQ$ are parallel. We know that $BMNC$ is a cyclic quadrilateral. Consequently, lines $BC$ and $MN$ are antiparallel with respect to lines $IB$ and $IC$. Therefore, $MN$ is parallel to the tangent at $I$ to the circumcircle of $\triangle IBC$, which implies that $MN \perp AI$. This result is the only element we will utilize from the original problem's proof. Define $w_m$ and $w_n$ to be the incircles of $\triangle ABD$ and $\triangle ACD$, respectively. Let $w_m$ and $w_n$ be tangent to $BC$ at points $T_m$ and $T_n$, respectively. Let the external tangent to $w_m$ and $w_n$, distinct from $BC$, intersect $AD$ at $E'$.

We now claim that $E$ and $E'$ are reflections of one another across $MN$. proof

Next, we claim that $AE'$ is equal to the length of the tangent drawn from $A$ to the incircle of $\triangle ABC$. proof

Finally, since $E$ and $E'$ are reflections about $MN$, we find that $EE' \perp MN$. Furthermore, $AQ \perp MN$. As $QE = AE'$ by construction, it follows that quadrilateral $AQEE'$ is a parallelogram. Therefore, $AD \parallel EQ$.

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ayeen_izady wrote: Let $K$ and $L$ be the intersection of $AP$ with $(CDP)$ and $(BDP)$ and let $I$ be the incenter of $ABC$. Note that $BP^2=BD\cdot BC$ thus $BP$ is tangent to $(CKDP)$ and similarly $CP$ is tangent to $(BDLP)$. $$\measuredangle PDK=180-\measuredangle PCK= 90+\measuredangle CPK=180-\measuredangle BPK=\measuredangle BDL=90+\measuredangle BDM=180-\measuredangle PDM$$So $M,D,K$ are collinear and similarly $N,L,D$ are collinear. $$\measuredangle AMI=90-\measuredangle BDM=\measuredangle PCK= \measuredangle BPL$$thus $APMB$ is cyclic and similarly $APNC$ is cyclic. So $IM\cdot IB=IP\cdot IA=IN\cdot IC$. Hence $BMNC$ is cyclic and we are done $\square$ [asy][asy] import geometry; size(8cm); pair A,B,C,P,D,M,N,K,L,I; A=dir(130); B=dir(210); C=dir(330);P=intersectionpoints(circle(B,C),line(A,incenter(A,B,C)))[1];D=foot(P,B,C);M=incenter(A,B,D);N=incenter(A,C,D);K=foot(C,A,P);L=foot(B,A,P); I=incenter(A,B,C); draw(A--B--C--cycle);filldraw((A--P--M--B--cycle),palered);filldraw((A--P--N--C--cycle),paleblue); draw(circle(P,D,C),royalblue);draw(circle(P,D,B),red);draw(A--K);draw(C--K); draw(B--L);draw(P--D);draw(M--I--N);draw(D--N);draw(M--K); dot(P);dot(B);dot(C);dot(A);dot(D);dot(M);dot(N);dot(K);dot(L);dot(I); label("$A$",A,dir(A));label("$B$",B,dir(B));label("$C$",C,dir(C));label("$D$",D,dir(240));label("$K$",K,dir(B));label("$P$",P,dir(80)); label("$M$",M,dir(130));label("$N$",N,dir(60));label("$L$",L,dir(160));label("$I$",I,dir(140)); [/asy][/asy] By saying that $\angle BDL=90^{\circ}+\angle BDM$, aren't you assuming that $D, L$ and $N$ are collinear? I mean, $\angle MDN=90^{\circ}$, that's clear, but you don't know if $\angle MDN = \angle MDL$.

surprisingly simple for a q4? headsolved while writing up it suffices to show that $BMPA$ is cyclic, as then analogously, $APNC$ is cyclic, which would suffice by radax at $I$. now, we just want $\angle AMB=90^\circ+\frac{\angle ADB}{2}=\angle APB$. Let $AP\cap BC$ be $X$. Let $Y$ be the harmonic conjugate of $X$ in $BC$. Now, we frame this in terms of $\triangle BPC$. $X$ is some arbitrary point on $BC$. Then, we construct the Apollonian circle of $A$ wrt $BC$ by drawing the semicircle through $X,Y$ and intersecting it with $PX$. Now, it is well-known by right angle harmonic lemma that $PB$ bisects $\angle YPX$. Further, $\angle YAX=90^\circ$ by Thales. Now, from opposite right angles, $AYDP$ is cyclic, so $\angle ADB=\angle APY$. Now, $\angle APB=90^\circ+\angle XPC=90^\circ+\frac{\angle ADB}{2}$ as desired, due to $PC$ being an external angle bisector in $\triangle XPY$. yay

sami1618 wrote: This was one of my favorite geometry problems . Here is a sketch of my solution to the extension: XX-math-XX wrote: Here's another interesting fact about this problem. Let $E$ be the tangency point of the incircle of triangle $ABC$ with $BC$, suppose that there is a point $Q$ on the segment $AI$ such that $EQ$ has the same length as the tangent drawn from $A$ to the incircle of triangle $ABC$. Prove that $AD$ and $EQ$ are parallel. We know that $BMNC$ is a cyclic quadrilateral. Consequently, lines $BC$ and $MN$ are antiparallel with respect to lines $IB$ and $IC$. Therefore, $MN$ is parallel to the tangent at $I$ to the circumcircle of $\triangle IBC$, which implies that $MN \perp AI$. This result is the only element we will utilize from the original problem's proof. Define $w_m$ and $w_n$ to be the incircles of $\triangle ABD$ and $\triangle ACD$, respectively. Let $w_m$ and $w_n$ be tangent to $BC$ at points $T_m$ and $T_n$, respectively. Let the external tangent to $w_m$ and $w_n$, distinct from $BC$, intersect $AD$ at $E'$.

We now claim that $E$ and $E'$ are reflections of one another across $MN$. proof

Next, we claim that $AE'$ is equal to the length of the tangent drawn from $A$ to the incircle of $\triangle ABC$. proof

Finally, since $E$ and $E'$ are reflections about $MN$, we find that $EE' \perp MN$. Furthermore, $AQ \perp MN$. As $QE = AE'$ by construction, it follows that quadrilateral $AQEE'$ is a parallelogram. Therefore, $AD \parallel EQ$. Let $(I)$ be the incircle of $\triangle ABC$ and let $(I)$ be tangent to $AB,AC$ at $U,V$ respectively (keep the previous notations) now $UM,VN,AD,IE$ are concurrent.

Solution using mostly angle chasing. Claim: We have $\angle BPD = \angle BAD + \angle ACP$ and $\angle DPC = \angle DAC + \angle PBA$. Proof: Let the reflection of the line $CA$ across $CP$ hit the $AB$ at $K$. Then $P$ is the incenter of triangle $\triangle AKC$. Thus, $\angle KPB = \angle KPC - 90^{\circ} = 90^{\circ} + \frac{\angle BAC}{2} - 90^{\circ} = \angle KAP$. Therefore $BK \cdot BA = BP^2 = BD \cdot BC$ so $AKDC$ is cyclic. Now, we have $$\angle BPD = \angle PCD = \angle PCK + \angle KCD = \angle BAD + \angle ACP$$The second result follows analogously. [asy][asy] size(13cm); import geometry; draw(unitcircle, blue+white); pair A = dir(130); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); circle c = circle((point)B, C); point[] P = intersectionpoints(c, line(A,I)); pair P = P[1]; pair D = intersectionpoint(line(B,C), perpendicular(P,line(B,C))); pair M = incenter(A,B,D); pair N = incenter(A,C,D); transform reflect = reflect(line(C,P)); pair K = intersectionpoint(line(A,B),reflect*line(A,C)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$P$", P, dir(P)); dot("$D$", D, dir(D)); dot("$I$", I, dir(I)); dot("$K$", K, dir(K)); draw(A -- B -- C -- cycle); draw(B -- P -- C); draw(A -- I); draw(P -- D -- A); draw(C -- K -- P); draw(circumcircle(A,K,D), red+dashed); markrightangle(B,P,C); markangle(A,C,P, purple); markangle(P,C,K,purple); markangle(K,P,B,green); markangle(B,A,P,green); [/asy][/asy] Now, note that $$\angle MAP = \angle BAP - \angle BAM = \angle BAC / 2- \angle BAD / 2 = \angle DAC / 2 $$On the other hand, if $P'$ is the isogonal conjugate of $P$, then $$\angle P'BP = \angle P'BA - \angle PBA = \angle CBP - \angle PBA = \angle DAC$$This implies that $$\angle MBP = \angle IBP = \angle P'BP / 2 = \angle DAC / 2$$Hence, we deduce that $\angle MAP = \angle MBP$, so $APMB$ is cyclic. Similarly, we get that $APNC$ is cyclic. Thus by PoP, $IM \cdot IB = IP \cdot IA = IN \cdot IC$, so $BMNC$ is cyclic as desired. [asy][asy] size(13cm); import geometry; draw(unitcircle, blue+white); pair A = dir(130); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); circle c = circle((point)B, C); point[] P = intersectionpoints(c, line(A,I)); pair P = P[1]; pair D = intersectionpoint(line(B,C), perpendicular(P,line(B,C))); pair M = incenter(A,B,D); pair N = incenter(A,C,D); transform reflect1 = reflect(line(B,I)); pair Pm = intersectionpoint(line(A,I), reflect1*line(B,P)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$P$", P, dir(P)); dot("$D$", D, dir(D)); dot("$I$", I, dir(I)); dot("$M$", M, dir(M)); dot("$N$", N, dir(N)); dot("$P'$", Pm, dir(Pm)); draw(A -- B -- C -- cycle); draw(B -- P -- C); draw(A -- Pm); draw(P -- D -- A); draw(B -- I -- C); draw(M -- A -- N); draw(B -- Pm -- C); markrightangle(B,P,C); draw(circumcircle(A,P,M),red+dashed); draw(circumcircle(A,P,C), red+dashed); [/asy][/asy]