Tangent Circles

[tanmoy]

We denote the circumcircle of a triangle $XYZ$ by $(XYZ)$.

Let $ABC$ be an acute angled triangle and $P$ be a point inside the triangle such that $\angle BPC=180^{\circ}−\angle A$.$BP,CP$ intersect $CA,AB$ at $E,F$.Circle $(AEF)$ intersects $(ABC)$ again at $G$.The circle with diameter $PG$ intersects $(ABC)$ again at $K$. $D$ is the projection of $P$ on $BC$ and $M$ is the midpoint of $BC$.Prove that the circle $(KPG)$ and circle $(KDM)$ are tangent to each other.

[Raiyan Jamil]

$Hints:$

Prove a lemma: $W,W'$ are two circles which are reflections of eachother under $BC$. Let $M$ be the midpoint of $BC$. Let $P,Q$ be two lines intersecting $W,W'$ at respectively such that $U,V,W,X$ lies on the same side of $BC$. Then $UVWX$ is cyclic.
Then, show that the centres of two respective circles and touchpoint(Here K) are collinear. Done...

[asif e elahi]

$Hints:$

Prove a lemma: $W,W'$ are two circles which are reflections of eachother under $BC$. Let $M$ be the midpoint of $BC$. Let $P,Q$ be two lines intersecting $W,W'$ at respectively such that $U,V,W,X$ lies on the same side of $BC$. Then $UVWX$ is cyclic.
Then, show that the centres of two respective circles and touchpoint(Here K) are collinear. Done...

Which two circles?

We denote the circumcircle of a triangle $XYZ$ by $(XYZ)$.

Let $ABC$ be an acute angled triangle and $P$ be a point inside the triangle such that $\angle BPC=180^{\circ}-\angle A$.$BP,CP$ intersect $CA,AB$ at $E,F$.Circle $(AEF)$ intersects $(ABC)$ again at $G$.The circle with diameter $PG$ intersects $(ABC)$ again at $K$. $D$ is the projection of $P$ on $BC$ and $M$ is the midpoint of $BC$.Prove that the circle $(KPG)$ and circle $(KDM)$ are tangent to each other.

Apply a negative inversion with center $P$ that sends $\bigodot ABC$ to itself.

[tanmoy]

This problem is an application of IMO 2015/3.

Claim 1:$G,P,M$ are collinear.
Proof:Let the extension of line $PD$ intersects arc $ABC$ of $(ABC)$ at $R$.Then obviously $P$ is the orthocenter of $\triangle RBC$.Let $GP \cap (ABC)=N$
Then $\angle NPC=\angle FPG=\angle FAG=\angle BAG=\angle BNG=\angle BNP$ $\Rightarrow$ $BN \parallel CP$.Samely,$BP \parallel CN$ $\Rightarrow$ $M$ is the midpont of segment $PN$ $\Rightarrow$ $P,M,N$ are collinear $\Rightarrow$ $G,P,M$ are collinear.

From Claim 1,$N$ is the reflection of $P$ over $M$.So,from a well known lemma,$N$ is the antipode of $R$ in $(ABC)$ $\Rightarrow$ $\angle PGR=90^{\circ}$

Now the problem is same as IMO 2015/3

[Raiyan Jamil]

Which two circles?

$KPG ,KDM$