$Coordinate$ $Bash$ let $BC$ be the $x-axis$ and altitude of $BC$ be the $y-axis$ let $C \equiv (c,0)$ $B \equiv (-b,0)$ $A \equiv (0,a)$ now let $P \equiv ( t_1,0)$ so $ Q \equiv (c-b-t_1,0)$ now equation of $AB$ and $AC$ ca be easily found out. so $ F \equiv (c-b-t_1, \frac{(c-t_1)a}{b})$ $ Q \equiv (t_1, \frac{(c-t_1)a}{c})$ now $H_1$ lies on $FQ$ and $H_2$ lies on $PE$ also $H_1$ can be found out by finding the equation of line through $P$ $\perp^r$ to $AB$ and then putting $x=c-b-t_1$ so $H_1 \equiv (c-b-t_1, \frac{(2t_1+b-c)b}{a})$ similarly $H_2 \equiv ( t_1, \frac{(2t_1+b-c)c}{a})$ so now slope of $H_1H_2= \frac{(c-b)}{a})$ now we find the equation of lines $PF$ and $QE$ to find $M$. so finding the equation of the line and substituting $ \frac{(c-t_1)a}{2t_1+b-c}=l $, we get equation of $PF$ is $lx+by-lt_1=0$ and that of $QE$ is $lx-cy+lb+lt-lc=0$ using these equations we find that $y=\frac{(2t_1+b-c)l}{b+c} = \frac{(c-t_1)a}{b+c}$ similarly $x=\frac{(b+t_1)(c-b)}{b+c}$ so $M \equiv ( \frac{(b+t_1)(c-b)}{b+c}, \frac{(c-t_1)a}{b+c} )$ now we see that slope of $AM$ is $- \frac{(c-b)}{a}= - \frac{1}{m_{H_1H_2}}$ thus proving that $AM \perp H_1H_2$

Cute. Let the parallel through $P$ to $QE$ meet $QF$ at $K$ and define $L$ similarly. Note that $BK \parallel AC$ and $CL \parallel AB$. Let the orthocentres of $BKP, CLQ$ be $T, U$. Then, $T, U$ are the orthology centres of $BFP, CEQ$, so $TU \perp AM$. But, $P(T, Q; E, H_1) = Q(U, P; F, H_2) \implies H_1H_2 \parallel TU \perp AM$. Alternatively, we can avoid orthology centre. Let $EF$ meet $BC$ at $V$. Easy to see, by similarity, that $V \in TU$. Let parallel through $V$ to $QE$ meet $PT, PF$ at $X_1, Y_2$ and parallel through $V$ to $AC$ meet $BT, BF$ at $Y_1, X_2$. Obviously, $X_1Y_1 \perp VT$ and $Y_2X_2 \parallel AM$. Note $P(X_1, Y_2; B, K) = B(Y_1, X_2; P, K)$ so $AM \parallel X_2Y_2 \parallel X_1Y_1 \perp VT$ and we are done.

Man, you have doubled the $M$...

Thank you, I think it is okay now.

Maybe i am too dumb and blind, but I think $\triangle BKP$ and $\triangle CLQ$ are not orthologic of centers $T$ and $U$.

My solution: Let the perpendicular from $ P $ to $ AQ $ cut $ QF $ at $ H_1^* $ . Let the perpendicular from $ Q $ to $ AP $ cut $ PE $ at $ H_2^* $ . Let $ D $ be the projection of $ A $ on $ BC $ and $ T=AM \cap BC $ . Let $ Y \in PE, Z \in QF $ such that $ DY \parallel QH_2^*, DZ \parallel PH_1^* $ . Easy to see $ H_1^*H_2^*YZ $ is a parallelogram . ... $ ( \star) $ Since $ (P,Q;B,T)=(F,MP \cap AB;B,A)=(QF \cap AC,E;C,A)=(Q,P;C,D) $ , so $ D, T $ are symmetry WRT the midpoint of $ BC \Longrightarrow AT $ is the isotomic conjugate of $ AD $ WRT $ \angle BAC $ . ... $ (\Psi) $ Since $ Rt \triangle PQH_1 \cup H_1^* \sim Rt \triangle ADB \cup Q $ and $ Rt \triangle QPH_2 \cup H_2^* \sim Rt \triangle ADC \cup P $ , so we get $ H_1H_1^*:PQ=BQ:AD=CP:AD=H_2H_2^*:PQ \Longrightarrow H_1H_1^*H_2^*H_2 $ is a parallelogram $ \Longrightarrow H_1H_2 \parallel H_1^*H_2^* $ , hence combine $ (\star) , (\Psi) $ and from the problem An extension of a problem of perpendicularity we get $ AM \perp H_1H_2 $ . Q.E.D

StanleyST wrote: Maybe i am too dumb and blind, but I think $\triangle BKP$ and $\triangle CLQ$ are not orthologic of centers $T$ and $U$. Typo, they are orthology centre of the triangles in question.

IDMasterz wrote: Then, $T, U$ are the orthology centres of $BFP, CEQ$, so $TU \perp AM$. Could you please give some references of this property ?

Dear Mathlinkers, ideas and shape of a synthetic proof 1. the symmetric of H1, H2 wrt the midpoint of BC are U, T 2. by symmetry, UT // H1H2 3. the triangles BFP and CEQ are orthologic (U, T orthologic centers) and perspective (perspective axis AM) , 4. according to the Sondat’s theorem, UT is perpendicular to AM and we are done… Sincerely Jean-Louis

Can anyone help me out on my bashy solution? I think it is along different lines to everyone else's here. With the help of GeoGebra, I have found that $AM$ is fixed. I trig-bashed (trust me, you do't want to see all the details) and got $$\frac{\sin\angle MAF}{\sin\angle MAE}=\frac{\sin B\tan B}{\sin C\tan C}=\text{constant},$$ but don't know where to go from there. Again from GeoGebra, I think the orientation of $\overline{H_1H_2}$ is also fixed, but I don't know how to prove it.

jlammy wrote: Can anyone help me out on my bashy solution? I think it is along different lines to everyone else's here. With the help of GeoGebra, I have found that $AM$ is fixed. I trig-bashed (trust me, you do't want to see all the details) and got $$\frac{\sin\angle MAF}{\sin\angle MAE}=\frac{\sin B\tan B}{\sin C\tan C}=\text{constant},$$but don't know where to go from there. Again from GeoGebra, I think the orientation of $\overline{H_1H_2}$ is also fixed, but I don't know how to prove it. jlammy, let AM meet BC at Z, and the pependicular from A to BC intersect BC at H. Then BZ=CH.

We consider $\triangle ABC$ as fixed and $P,Q$ varying. Claim 1: Direction of line $H_1H_2$ is fixed (in other words, all such lines $H_1H_2$ as $P,Q$ vary are parallel). Proof: Since $\triangle PQH_1 \stackrel{+}{\sim} FQB$ and $QPH_2 \stackrel{+}{\sim} EPC$, so shape of $\triangle PQH_1,\triangle PQH_2$ are fixed. It follows shape of quadrilateral $H_1QPH_2$ is fixed (in other words, all such quadrilaterals $H_1QPH_2$ are directly similar). So $\angle (H_1H_2,PQ)$ is fixed. Since $PQ$ is always parallel to $BC$, hence direction of line $H_1H_2$ is fixed. $\square$ Claim 2: Line $AM$ is fixed. Proof: Let $D$ be foot of perpendicular from $A$ onto $BC$ and $N = AM \cap BC$. We will show $N,D$ are isotonic wrt $BC$. This shows $N$ is fixed, hence $AM \equiv AN$ is fixed. By DDIT on lines $PE,QF,PF,QE$ through point $A$ gives pairs $$ (AE,AF),(AP,AQ),(A \infty_{AD}, AM) $$swap under an involution. Project this onto $BC$. We get an involution swaps $$ (C,B),(P,Q),(D,N) $$As reflection in midpoint of $BC$ swaps first two pairs, so it must swap other pair also, as desired. $\square$ Hence it suffices to solve our problem for just one choice of $P,Q$! Note when $AB = AC$, our problem is easily true by symmetry. Assume $AB \ne AC$. Consider the case $Q=D$ and $P=N$. We want $B,H_1,H_2$ collinear in this case. An easy bash gives $$ \frac{BQ}{BP} = \frac{H_1Q}{H_2P} = \frac{\cot B}{\cot C}$$Note we need $AB \ne AC$, otherwise points $P,Q,H_1,H_2$ coincide, and line $H_1H_2$ doesn't make sense. This completes the proof. $\blacksquare$