In a triangle $\triangle ABC$ with orthocenter $H$, let $BH$ and $CH$ intersect $AC$ and $AB$ at $E$ and $F$, respectively. If the tangent line to the circumcircle of $\triangle ABC$ passing through $A$ intersects $BC$ at $P$, $M$ is the midpoint of $AH$, and $EF$ intersects $BC$ at $G$, then prove that $PM$ is parallel to $GH$. Proposed by Sreejato Bhattacharya
Problem
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Tags: geometry, circumcircle, trigonometry, power of a point, radical axis, exterior angle
sayantanchakraborty
22.12.2014 06:35
This problem can be killed by just a few lines of trigonometry,but since this is my friend's problem,I shall solve it in a nice way. Let $M_a$ be the midpoint of $BC$.$FE$ is the radical axis of the nine-point circle and that of $BCEF$,while $BC$ is the radical axis of the nine-point circle and $ADM_a$ where $D=AH \cap BC$.Thus $G$ lies on the radical axis of $ADM_a$ and $BCEF$.Also note that power of $H$ w.r.t both the circles is same.This enables us to conclude that $GH \perp AM_a$ since $AM_a$ is the line joining centres. $BC$ is the radical axis of $BCEF$ and $ABC$ while the tangent through $A$ is the radical axis of $ABC$ and the point circle $A$.Thus $P$ lies on the radical axis of $A$ and $BCEF$.We also have $ME=MF=MA$ and also note that $\angle{MEF}=\angle{ECF}=90-A$ so $M$ lies on their radical axis as well.Thus $PM$ is perpendicular to their centre joining line $AM_a$. Thus $PM \parallel GH$.
AnonymousBunny
22.12.2014 20:39
^ Yep, that's moreorless the intended solution... Unfortunately, there exists a way simpler and thoughtless solution using trig.
DaChickenInc
10.01.2015 00:12
Fortunately, trig and radical axes are both not required.
Define $D$ to be the intersection of $AH$ and $BC$ and $Q$ to be the intersection of $DF$ and $AP$.
Since $F$ is on $AB$, $\angle FAP=\angle BAP$. Since $AP$ is tangent to the circumcircle of $\triangle ABC$, $\angle BAP=\angle ACB$. $E$ is defined to lie on $AC$, so $\angle ACB=\angle BCE$. Furthermore, $\angle BFC=90^\circ=\angle BEC$, $BCEF$ is cyclic. $AF$ intersects the circumcircle of $BCEF$ again at $B$, so $\angle BCE=\angle AFE$. Combining equalities gives us $\angle FAP=\angle AFE$, so $AP\parallel EF$.
Consider the circle with center $M$ that passes through $A$. Since $M$ is the midpoint of $AH$, $AH$ is a diameter of the circle. Since $\angle AFH=90^\circ$, $F$ is on this circle, so $MA=MF$. $\angle AFQ$ is the exterior angle of $\triangle ADF$, so $\angle AFQ=\angle ADF+\angle DAF$. $B$ lies on $AF$ and $C$ lies on $BD$, so $\angle DAF=\angle DAB=90^\circ-\angle ABD=90^\circ-\angle ABC$. Since $H$ lies on $AD$, $\angle ADF=\angle HDF$. Because $\angle BFH+\angle BDH=90^\circ+90^\circ=180^\circ$, $BFHD$ is cyclic, which tells us that $\angle HDF=\angle HBF$. Since $A,E,C$ lie on $BF,BH,AE$, respectively, $\angle HBF=\angle EBA=90^\circ-\angle BAE=90^\circ-\angle BAC$. Putting this together, $\angle AFQ=90^\circ-\angle BAC+90^\circ-\angle ABC=180^\circ-\angle BAC-\angle ABC=\angle ACB$. Recall from tangents that $\angle FAP=\angle ACB$, so $\angle FAQ=\angle ACB=\angle AFQ$. Thus $\triangle AFQ$ is isosceles with $QA=QF$.
Since $MA=MF$, $QA=QF$, and $MQ=MQ$, SSS congruence tells us that $\triangle MAQ\cong\triangle MFQ$, so we can write $\angle FQM=\frac{2\angle FQM}{2}=\frac{\angle FQM+\angle AQM}{2}=\frac{\angle AQF}{2}=\frac{180^\circ-\angle FAQ-\angle AFQ}{2}=\frac{180^\circ-2\angle ACB}{2}=90^\circ-\angle ACB$. Also, since $C$ lies on $FH$, $\angle DFH=\angle DFC$. Furthermore $\angle ADC=90^\circ=\angle AFC$, so $ACDF$ is cyclic, and $\angle DFC=\angle DAC=90^\circ-\angle ACB=\angle FQM$. But $D$ lies on $FQ$, so $\angle DFH=\angle FQM=\angle DQM$. Collinearity also tells us that $\angle FDH=\angle QDM$, so by AA similarity, $\triangle DFH\sim\triangle DQM$, which gives $\frac{DQ}{DF}=\frac{QM}{FH}$.
Since $AP\parallel EF$, collinearity tells us that $PQ\parallel GF$, and $D$ lies on $PG$ and $QF$, $\angle DGF=\angle DPQ$ and $\angle DFG=\angle DQP$, so AA similarity gives us $\triangle DFG\sim\triangle DQP$. As a result, $\frac{PQ}{FG}=\frac{DQ}{DF}=\frac{QM}{FH}$. Furthermore, $\angle PQM=\angle PQD+\angle DQM=\angle DFG+\angle DFH=\angle HFG$, so by SAS similarity, $\triangle FGH\sim\triangle QPM$. This, along with our previous similarity, tells us that $\angle DPM=\angle DPQ-\angle QPM=\angle DGF-\angle FGH=\angle DGH$. Since $D,P,G$ lie on the same line, $PM\parallel GH$.
Sorry for lack of motivation.
TheOverlord
16.01.2015 18:40
Another solution: Let $D$ be the foot of $A$-perpendicular and $Q$ the intersection of $AD$ and $EF$. Obviously $P \hat{A} D=\hat{C}=A\hat{F}E \Rightarrow AP||QG\Rightarrow \frac{DQ}{DA}=\frac{DG}{DP}$. We must prove $GH||PM \Leftrightarrow \frac{DH}{DM}=\frac{DG}{DP}=\frac{DQ}{DA} \Leftrightarrow DB.DC=DH.DA=DM.DQ $. But $AEFH$ is concyclic quadrilateral with circumcentre $M$, and $B=HE\cap AF$, $C=HF\cap AE$ and $Q=EF\cap AH$. So by Brokard's theorem ,$M$ is the orthocenter of $BQC$, and $D$ is the foot of $Q$-perpendicular, so $DP.DM=DB.DC$. Q.E.D.