My solution : Let $ D= \odot (I) \cap BC $ and $ T $ be the reflection of $ D $ in $ H $ . Let $ Q', R' $ be the projection of $ Q, R $ on $ AB, AC $, respectively . Let $ M $ be the midpoint of $ AH $ and $ Y=BI \cap \odot (AEF), Z=CI \cap \odot (AEF) $ It's well-known $ Y, M, Z $ are collinear at A-midline of $ \triangle ABC $ . From symmetry $ \Longrightarrow \triangle QQ'F \sim \triangle QHD \sim \triangle QHT $ , so $ \angle HTQ=\angle Q'FQ \Longrightarrow T \in \odot (BFQ) $ ( similarly, $ T \in \odot (CER) $ ) , hence from Miquel theorem we get $ G =\odot (BFQ) \cap \odot (CER) $ ( $ G \neq T $ ) lie on $ \odot (AEF) $ . From Mannheim theorem $ \Longrightarrow G \in \odot (IQR) $ , so $ \angle MQG=\angle ZIG=\angle MYG \Longrightarrow G, Q, M, Y $ are concyclic , hence from $ \angle TGQ=\angle TBQ=\angle MYQ=180^{\circ}-\angle QGM $ we get $ M \in GT $ . i.e. the radical axis $ TG $ of $ \odot (BFQ) $ and $ \odot (CER) $ pass through the midpoint $ M $ of $ AH $ Q.E.D

TelvCohl wrote: From Mannheim theorem Sorry What is Mannheim Theorem

ThE-dArK-lOrD wrote: TelvCohl wrote: From Mannheim theorem Sorry What is Mannheim Theorem I think that is Mannheim

My solution: $D \equiv \odot (I) \cap BC$, let $T$ be the reflection of $D$ in $H$, $\angle REA=\angle RDT=\angle RTD \Rightarrow T \in \odot (REC)$, analogously $T \in \odot (QFB)$, $G$ is the Miquel point of $\triangle ABC$ WRT $T,E,F$, the parallel line from $A$ to $BC$ cut again to $\odot (EFGI)$ at $X \Rightarrow AHDX$ is a rectangle, and the midpoint of $AH$ belongs to $TX$. $\angle BCA=\angle EAX=\angle XGE \Rightarrow T,G,X$ are collinear $\Rightarrow TX$ is the radical axis between $\odot (REC)$ and $\odot (QFB)$

Here’s my solution: Let $M$ be the midpoint of $AH$. $D$$\equiv$$\odot{(I)}$$\cap$$BC$. $J$ is the reflection of $D$ WRT midpoint of $BC$, then $JI$ passes through $M$ (easy to recognize). $DF$$\cap$$AH$$\equiv$$K$, $DE$$\cap$$AH$$\equiv$$L$. It is well known that $JK\parallel{CI}$ and $JL\parallel{BI}$ $\Rightarrow$ $H_M$: $I$$\longrightarrow$$J$, $R$$\longrightarrow$$K$, $Q$$\longrightarrow$$L$ $\Rightarrow$ $\frac{MR}{MQ}$=$\frac{MK}{ML}$ $\Rightarrow$ $MR.ML$=$MQ.MK$ (1) On the other hand: $DL\perp{CI}$, $CD\perp{HL}$ $\Rightarrow$ $\angle{RLE}$=$\angle{DCI}$ $\Rightarrow$ $\angle{RLE}$=$\angle{RCE}$ $\Rightarrow$ $R, E, C, L$ are on the circle (2) Analogously, we have $K, F, B, Q$ are on the circle (3) (1), (2), (3) $\Rightarrow$ $M$ has the same power WRT $\odot{(REC)}$ and $\odot{(QFB)}$, and the conclusion follows. PS. This is really a nice problem.

Here is an generalization Let $ABC$ be a triangle. Incircle $(I)$ touches $CA,AB$ at $E,F$. $P,Q$ lie on $IC,IB$ such that $PQ$ is perpendicular to $BC$ at $H$. $K$ is projection of $A$ on $PQ$. Prove that radical axis of circles $(PEC),(QFB)$ bisects $HK$.

buratinogigle wrote: Here is an generalization Let $ABC$ be a triangle. Incircle $(I)$ touches $CA,AB$ at $E,F$. $P,Q$ lie on $IC,IB$ such that $PQ$ is perpendicular to $BC$ at $H$. $K$ is projection of $A$ on $PQ$. Prove that radical axis of circles $(PEC),(QFB)$ bisects $HK$. My proof for the original problem ( post #2 ) also works for this generalization

tranquanghuy7198 wrote: Here’s my solution: It is well known that $JK\parallel{CI}$ I think that this is not well-known. Maybe you means to this: If the $A$-excircle touch $AC$ at $X$, then $K,J,X$ are collinear $\Rightarrow KJ \parallel CI$ Original problem The incircle of $\triangle ABC$ touch $BC,CA$ and $AB$ at $D,E$ and $F$ respectively. The $A$-excircle touch $BC,CA$ and $AB$ at $D_1,E_1$ and $F_1$ respectively. $K \equiv FD \cap E_1D_1$ Prove that $AK \perp BC$

buratinogigle wrote: Here is an generalization Let $ABC$ be a triangle. Incircle $(I)$ touches $CA,AB$ at $E,F$. $P,Q$ lie on $IC,IB$ such that $PQ$ is perpendicular to $BC$ at $H$. $K$ is projection of $A$ on $PQ$. Prove that radical axis of circles $(PEC),(QFB)$ bisects $HK$. My solution (post #5) imply this, too

You are right, drmzjoseph, I want to mention about Paul Yiu's theorem, which is very famous. It can be read with more detail in mathscope forum.

Fortunately, my solution in post #6 still works, too, but there is a bit different. We cannot use Paul Yiu’s theorem any more, we must use the extension of it: Extension. Let the incircle $(I)$ of $\triangle{ABC}$ touch $BC, CA, AB$ at $D, E, F$, resp. $K$ is the point such that $AK\parallel{BC}$, $H$ is the projection of $K$ on $BC$, $HK$$\cap$$DF$$\equiv$$L$. $DG$ is a diameter of $\odot{(I)}$, $KG$$\cap$$BC$$\equiv$$J$. Prove that $JL\parallel{CI}$ Prove. Other points are constructed as in the figure. Paul Yiu tells us that $J’L’\parallel{CI}$. On the other hand, we have: $\frac{DL}{DL’}$ = $\frac{MK}{MA}$ = $\frac{DJ}{DJ’}$ $\Rightarrow$ $JL\parallel{J’L’}$ and the conclusion follows. Everything other than the above stays the same as my post #6

mathscope forum? What's that? When I say this: If the $A$-excircle touch $AC$ at $X$, then $K,J,X$ are collinear $\Rightarrow KJ \parallel CI$, I mentioned "the original" problem, it's just the picture, (by extraversion), this is well-known: [asy][asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -31.827963549562938, xmax = 34.69829095715714, ymin = -4.918376404737516, ymax = 36.029486824525364; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((6.627564729864586,0.7994712465383367)--(6.643463779672261,3.1792410217509777)--(4.263694004459619,3.1951400715586518)--(4.247794954651945,0.8153702963460111)--cycle, qqwuqq); /* draw figures */ draw(circle((19.184392850360236,12.8930035247441), 12.177151674855965)); draw(circle((-13.549713948117091,15.60392208427459), 14.66932069072253)); draw((-13.647716308629148,0.9349287628297205)--(19.10304009834057,0.7161236024410867)); draw((12.189714581818947,0.7623109722735415)--(-0.9032435403441017,23.037343621667862)); draw((4.458380885632265,32.33587244548175)--(-13.647716308629148,0.9349287628297205)); draw((4.458380885632265,32.33587244548175)--(19.10304009834057,0.7161236024410867)); draw((9.844534269753431,20.706457601316945)--(-6.734390792107515,0.8887413929972656)); draw((4.458380885632265,32.33587244548175)--(4.247794954651945,0.8153702963460111), linetype("2 2")); /* dots and labels */ dot((4.336699715932933,14.122634964884837),dotstyle); label("$A$", (4.969020393445808,13.928859273388959), NE * labelscalefactor); dot((-6.734390792107515,0.8887413929972656),dotstyle); label("$B$", (-7.483617465316298,-1.5527986050720735), NE * labelscalefactor); dot((12.189714581818947,0.7623109722735415),dotstyle); label("$C$", (11.81236191943219,-1.4406126784165587), NE * labelscalefactor); dot((-13.647716308629148,0.9349287628297205),linewidth(2.pt) + dotstyle); dot((-0.9032435403441017,23.037343621667862),linewidth(2.pt) + dotstyle); dot((19.10304009834057,0.7161236024410867),linewidth(2.pt) + dotstyle); dot((9.844534269753431,20.706457601316945),linewidth(2.pt) + dotstyle); dot((4.458380885632265,32.33587244548175),linewidth(2.pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

I understand, drmzjoseph, that was exactly what I wanted to say!!!

Nice problem!

Here is my solution for this problem Solution Let $D$ be tangency point of ($I$) with $BC$; $M$ be midpoint of $AH$; $X$ $\equiv$ $DF$ $\cap$ $AH$; $Y$ $\equiv$ $DE$ $\cap$ $AH$ We have: ($XH$; $XD$) $\equiv$ ($DI$; $DX$) $\equiv$ ($BQ$; $BF$) (mod $\pi$) or $B$, $Q$, $X$, $F$ lie on a circle Similarly: $C$, $R$, $Y$, $E$ lie on a circle We also have: $\dfrac{\overline{MR}}{\overline{MQ}}$ = $\dfrac{\overline{MX}}{\overline{MY}}$ Then: $\overline{MR}$ . $\overline{MY}$ = $\overline{MQ}$ . $\overline{MX}$ or $P_{M / (REC)}$ = $P_{M / (BQF)}$ Hence: $M$ lies on radical axis of ($REC$) and ($BQF$) For the proof of: $\dfrac{\overline{MR}}{\overline{MQ}}$ = $\dfrac{\overline{MX}}{\overline{MY}}$, you can see here

Here's a solution with Linearity of PoP WLOG $AB < AC.$ Let $X$ be the point where the incircle touches $BC$, and let $X'$ be the reflection of $X$ over $H.$ Lemma. $X'$ is on $(\triangle KPB).$ Proof. We have that $\angle PX'X = \angle PXX' = \angle KXB = \angle BKX = \angle BKP$, which implies the result. $\blacksquare$ From Linearity of PoP, it suffices to prove that $AK \cdot AB - AL \cdot AC = -(HX' \cdot HB + HX' \cdot HC)$, i.e. we want $AK(AC - AB) = HX' \cdot BC.$ This is a rather straightforward computation after remembering that $HX = HX'$. $\square$

The same problem with different labeling wrote: The inscribed circle of an acute $\Delta ABC$ is tangent to $AB$ and $AC$ in $K$ and $L$ respectively. The altitude $AH$ intersects the angle bisectors of $\angle ABC$ and $\angle ACB$ in $P$ and $Q$ respectively. Prove that the middle point $M$ of $AH$ lies on the radical axis of the circumscribed circles of $\Delta KPB$ and $\Delta LQC$. Wlog $AB<AC$. Let $D$ be the intersection of $(QPI)$ and $(AKIL)$. Claim. $D$ lies on $(BKP)$ and $(QLC)$. \begin{align*} \measuredangle KDP&=\measuredangle IDP+\measuredangle KDI\\&=\measuredangle IQP+\measuredangle KAI\\&=0.5\measuredangle ACB+0.5\measuredangle BAC- 90^\circ\\&=-0.5\measuredangle CBA\\&=\measuredangle KBP\end{align*}We do the same for $(QLC)$. The claim follows. By Miquel's theorem, $E=(KDPB)\cap (CLQD)$ lies on $BC$. Let $N$ be intersection of $(I)$ and $BC$. Claim. $N$ is the reflection of $E$ over $AH$. Note that as $\measuredangle EBP=\measuredangle PBK$, we have $PE=PK$. Also, as $BK=BK$, by the congruent triangles $\triangle NBP\cong \triangle KBP$, hence $PN=PK$ and therefore $N$ is indeed the reflection of $E$ over $AH$. Thus, $HE=HN$. We use linearity of pop now. \begin{align*}\mathbb{P}(M,(KDPB),(CLQD))&=\frac{1}{2}\mathbb{P}(A,(KDPB),(CLQD))+\frac{1}{2}\mathbb{P}(H,(KDPB),(CLQD))\\ &=\frac{1}{2}(AK\cdot AB-AL\cdot AC+HE\cdot HB+HE\cdot BC) \\ &=\frac{1}{2}(AK\cdot (AB-AC)+HN\cdot BC) \end{align*}Claim. $AK\cdot (AC-AB)=HN\cdot BC$ Let $X$ reflection of $B$ over $AI$, it is well-known that $X$ lies on $(BIC)$. $$\frac{AC-AB}{BC}=\frac{\sin{\angle CBX}}{\sin{\angle BIC}}$$and $$\frac{HN}{AK}=\frac{HN}{\sin{\angle BIC}\cdot AI}=\frac{\sin{\angle HAI}}{\sin{\angle BIC}}.$$By angle chase, we actually have \begin{align*}\measuredangle HAI&=0.5\measuredangle BAC-\measuredangle BAH\\&= 0.5\measuredangle BAC+\measuredangle CBA-90^\circ\\&=\measuredangle CBA- \measuredangle XBA=\measuredangle CBX \end{align*}Therefore, in fact the claim is true. Hence, $\mathbb{P}(M,(KDPB),(CLQD))=0$ and we are done.

drmzjoseph wrote: The incircle $\odot (I)$ of $\triangle ABC$ touch $AC$ and $AB$ at $E$ and $F$ respectively. Let $H$ be the foot of the altitude from $A$, if $R \equiv IC \cap AH, \ \ Q \equiv BI \cap AH$ prove that the midpoint of $AH$ lies on the radical axis between $\odot (REC)$ and $\odot (QFB)$ I hope that this is not repost Let $M$ be the midpoint of $AH$ and $DI$ cut $(AI)$ at $J$. Let $P$ be the reflection point of $D$ about $H$. Since $\angle RPD=\angle RDP =\angle REA$, we get $R,E,C,P$ are concyclic. Similarly, $Q,F,B,P$ are concyclic, so if $(REC)$ cuts $(QFB)$ at $L$ then $L\in (AI)$. Let $AI$ cut $BC$ at $N$, then a simple angle chasing yields $P,L,I,N$ are concyclic, but $AJ\parallel BC$, so by Reim's, $P,L,J$ are collinear. Combining with $P,M,J$ are clearly collinear, we are done.