it is a very easy question if you know polars and a hard one if you do not know let $CC'\perp AB$ so the cercle around the point $B,A,Q,P$ have the center at $M$ so $C'$ is the harmonical conjugate of $T$ versus the circle so $C$ is on the polar of $T$ using the duality principle we get that $T$ is on the polar of $C$ so $TH\perp CM$ as $M$ is the center

another nice solution using the notes from first solution we now take $K$ on $CM$ such that $HK\perp CM$ so now $C'MHK,QPHKC,QPC'M$ are cyclic the last is cyclic cose all 4 point are on the Eulres circle but from radical axis or Monge theorem we get that $QP,C'M,HK$ are concurent but $QP,C'M$ are concurent at $T$ so we get that $HT\perp CM$ so we are done

Even though the statement seems nice, this problem can be trivially solved using Brockard's Theorem... Since $\angle APB=\angle AQB=90^{\circ}$, $A,B,P,Q$ are concyclic, so by Brockard's Theorem $M$ is the orthocentre of $CHT$, therefore $CM\perp TH$.

Just harmonic division . It was also Brazil TST (2001 or 2002)

Let $K$ be the perpendicular from $H$ to $CM$, let $CH$ meet $AB$ at $N$. Then as $\angle HNM= \angle HKM=90$, $KHNM$ is cyclic. Also, as $\angle HKC=\angle HQC= \angle HPC=90$, $KHPCQ$ is cyclic. Finally, as $P$,$Q$,$M$, and $N$ lie on the Nine-point circle, $PQMN$ is cyclic. By the radical axis theorem, the radical axes of these three circles, $PQ$, $MN$, and $KH$, must concur, so $KH$ passes through $T$. So $TH \perp CM$. Many other olympiad problems where we wish to prove perpendicularity from an outside point, can be solved similarly, by dropping a perpendicular from the middle of the three points to the desired line, finding three cyclic quadrilaterals (often using the nine-point circle), and then using the radical axis theorem. I've seen it in three or four different problems. EDIT: I now see that this is the same solution as frenchy's but in more detail. But I see he should have written $C'$ is the perpendicular from $C$ to $AB$ rather than $AC$, that was why I was confused.

Old problem ... Very nice and well-known the frenchy's or dgreenb801's proof without polars or proiective geometry. I"ll come back soon with another proof.

Suppose $a,b,p,q$ is on the unit circle of a complex plane. We can see that $m=0$, $h=\dfrac{ap(b+q)-bq(a+p)}{ap-bq}$, $t=\dfrac{ab(p+q)-pq(a+b)}{ab-pq}$ and $c=\dfrac{aq(b+p)-bp(a+q)}{aq-bp}$. also $\overline{t}=\dfrac{b+q-a-p}{bq-ap}$, $\overline{t}=\dfrac{p+q-a-b}{pq-ab}$, and ${\overline c}=\dfrac{b+p-a-q}{bp-aq}$. we need to prove that $\dfrac{t-h}{\overline{t}-\overline{h}}=-\dfrac{c-m}{\overline{c}-\overline{m}}$. the computation is messy but at least you know where it will solve the problem. $m=0$ is very helpful though.

See here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46146 ...

Generalization. Let $ABC$ be a triangle. For a point $P$ consider the points $D\in BC$ , $E\in CA$ , $F\in AB$ so that $BFPD$ , $CDPE$ are cyclically. Denote $m(\angle PDC)=m(\angle PEA)=m(\angle PFB)=\phi$ and $L\in EF\cap BC$ . The sideline $BC$ cut again the circumcircle of $\triangle DEF$ in the point $K$ . Prove that $m\left(\widehat {LP,AK}\right)\in \{\phi ,\pi -\phi\}$ . Particular cases. PC1. Let $ABC$ be a triangle. For a point $P$ denote its projections $D\in BC$ , $E\in CA$ , $F\in AB$ on the sidelines of $\triangle ABC$ . Denote $L\in EF\cap BC$ . The sideline $BC$ cut again the circumcircle of $\triangle DEF$ in the point $K$ . Prove that $LP\perp AK$ . PC2. Let $\triangle ABC$ . Denote the midpoint $M$ of the side $[BC]$ , the projections $E$ , $F$ of the orthocenter $H$ to the sidelines $AC$ , $AB$ respectively and the intersection $L\in BC\cap EF$. Prove that $LH\perp AM$ . PC3. Let $ABC$ be a triangle with incircle $C(I)$ which touches the sides of $\triangle ABC$ in $D\in BC$ , $E\in CA$ , $F\in AB$ . Denote $T\in BC\cap EF$ . Prove that $TI\perp AD$ . PC4. Let $ABC$ be a triangle with exincircle $C(I_a)$ which touches the sides of $\triangle ABC$ in $D'\in BC$ , $E'\in CA$ , $F'\in AB$ . Denote $T'\in BC\cap E'F'$ . Prove that $T'I_a\perp AD'$ .

Nice generalization, dear Virgil Circles $\odot(BDF),\odot(CDE),\odot(AFE)$ concur at their Miquel point $P$ WRT $\triangle ABC.$ Let $U \equiv AK \cap LP$ and $V,U'$ the second intersections of $\odot(AFE)$ with $LA,LP,$ respectively. Then $P$ becomes Miquel point of $\odot(AVE),$ $\odot(LDV),$ $\odot(CED)$ WRT $\triangle ALC,$ in other words, $P,V,L,D$ are concyclic. In the inversion through pole $L$ with power $LP \cdot LU',$ we have that $V \mapsto A,$ $U' \mapsto P$ and $K \mapsto D$ $\Longrightarrow$ points $A,U',K$ are collinear $\Longrightarrow$ $U \equiv U'.$ Then $\angle(LP,KA)=\angle(FP,FA)=\phi \ \pmod\pi.$

Akashnil wrote: Let $ABC$ be a triangle in which $BC<AC$. Let $M$ be the mid-point of $AB$; $AP$ be the altitude from $A$ on $BC$; and $BQ$ be the altitude from $B$ on to $AC$. Suppose $QP$ produced meets $AB$ (extended) in $T$. If $H$ is the ortho-center of $ABC$, prove that $TH$ is perpendicular to $CM$. A different solution (OWN) can be found here

Let $ABC$ be a triangle in which $AC<AC$. Let $M$ be the mid-point of $BC$, $CF$ be the altitude from $C$ on $AB$, and $BE$ be the altitude from $B$ on to $AC$. Suppose that $EF$ produced meets $BC$ (extended) at $T$. If $H$ is the orthocenter of $ABC$, prove that $TH$ is perpendicular to $CM$. I did not read the other solutions. So I don't know if it has been posted before. Clearly, $ (TBDC)=-1 $ From this easy to see that $ TD.TM=TB.TC $ Consider the circles $ \odot ADM $ and $ \odot BEFC $. Both of their centres lie on $ AM $. $ H $ lie on their radical axis since $ BH.HE=AH.HD $ $ T $ lies on the radical axis too because $ TD.TM=TB.TC $ So $ TH \perp AM $

Another nice solution using the following lemma: Lemma: Let $ABCD$ is a quadrilateral inscribed in a circle centred at O. If $AB\cap DC=R, AD\cap BC=Q, AC\cap BD=P$, then $ OP\perp QR $.(Also, $OP\cap QR$ is the Miquel point of $ABCD$). Applying this lemma to the problem, $MH\perp TC$, therefore $H$ is the orthocentre of $ \triangle TCM $ and the result follows

sax wrote: Another nice solution using the following lemma: Lemma: Let $ABCD$ is a quadrilateral inscribed in a circle centred at O. If $AB\cap DC=R, AD\cap BC=Q, AC\cap BD=P$, then $ OP\perp QR $.(Also, $OP\cap QR$ is the Miquel point of $ABCD$). can you post a proof for your lemma?

See this http://web.mit.edu/yufeiz/www/olympiad/cyclic_quad.pdf. It is the fact 15..

sax wrote: See this http://web.mit.edu/yufeiz/www/olympiad/cyclic_quad.pdf. It is the fact 15.. thanks, a very interesting and useful article.do you have more such articles?

$tan(C-x)=\frac{BE}{HP}=\frac{BP}{HP}.(\frac{tan(B)}{tan(C)+2tan(B)})$ , hence $tan(C-x)=\frac{tan(C)tan(B)}{tan(C)+2tan(B)}$ and so by (*) we have $x=y$ , done!

Another nice solution: Let $X$ be the foot of altitude from $C$ to $AB$.Then Radical axis(nine point circle,$\odot{CXM}$)=$AB$. Radical axis(nine point circle,$\odot{BPQA}$)=$PQ$. As $PQ \cap AB = T$ we get that $T$ is a point on the radical axis of $\odot{CXM},\odot{BPQA}$. As $HC*HX=HP*AH$,$H$ is also a point on the radical axis of $\odot{CXM},\odot{BPQA}$. $TH$ is the radical axis of the two circles. As the center of $\odot{CXM}$ lies on $CM$ and the center of $\odot{ABPQ}$ is $M$,and radical axis of two circles is perpendicular to the line joining centres,we obtain $TH\perp CM$. Bye... Sayantan....

$ABPQ$ is $Concyclic$ with center at $M$ , radius $= \frac{AB}{2} $. Denote this circle by $ \omega$. $AB \cap PQ = T$. $AQ \cap PB = C$. $AP \cap QB = H$. So , $TCH$ is self polar triangle with center of $ \omega$ as its orthocentre. As $M$ is center of $ \omega$ , $TH\perp CM$.

Use Brocard's theorem and the problem was solved

The point of intersection of the perpendicular lines is well known as the vertex of the D-triangle of $ABC$ corresponding to $C$ (Roger Johnson), HM-point (article in MR), and the C-Humpty point (here and here.) Just for documentation.

MY SOLUTION Brocard theorem on $(ABPQ)$$\implies TH\perp CM$

I proved this as a Lemma in solving 2016 USA TSTST P2 about a month ago lol

Applying Brocard theorem for cyclic quadrilateral $ABPQ$, we have: $H$ is orthocenter of $\triangle$ $CMT$ or $HT$ $\perp$ $MC$

By Brokard theorem in $(ABPQ)$ where the center is $M$. $\implies \triangle THC$ is autopolar. $\implies CM \perp TH$

Nothing new. Since $\measuredangle APB=\measuredangle AQB=90^\circ$, points $A,B,P,Q$ lie on a circle with center $M$. By Brokard's Theorem, $M$ is the orthocenter of $\triangle CHT$ which completes the proof.

Applying Brokard's Theorem on $ABPQ$, we have that $M$ is the orthocenter of $\triangle CHT$ so we take the opportunity to practice compleks instead. We set the circle centered at $M$ with radius $MA$ be the unit circle. We know that this circle passes through the points $B$ , $P$ and $Q$. WLOG rotate so that $a=1$ and thus $b=-1$. Now, using the chord intersection formula, we calculate \begin{align*} t &= \frac{ab(p+q)-pq(a+b)}{ab-pq}\\ &= \frac{p+q}{pq+1} \end{align*}\begin{align*} c &= \frac{aq(b+p)-bp(a+q)}{aq-bp}\\ &= \frac{2pq+p-q}{p+q} \end{align*}and \begin{align*} h &= \frac{ap(b+p)}{aq-bp}\\ &= \frac{2pq +q-p}{p+q} \end{align*} Now, we simply note that \begin{align*} \frac{c-m}{t-h} &= \frac{\frac{2pq+p-q}{p+q} - 0}{\frac{p+q}{pq+1} - \frac{2pq +q-p}{p+q}}\\ &= \frac{(2pq+p-q)(pq+1)}{p^2+q^2+p-q+p^2q-pq^2-2p^2q^2}\\ &= \frac{(2pq+p-q)(pq+1)}{-(2pq+p-q)(pq-p+q-1)}\\ &= - \frac{pq+1}{pq-p+q-1} \end{align*}Now the path is clear. Simply note that \begin{align*} \overline{\left(\frac{c-m}{t-h}\right)} &= \overline{\left(- \frac{pq+1}{pq-p+q-1}\right)}\\ &= -\frac{\frac{1}{pq}+1}{\frac{1}{pq}-\frac{1}{p}+\frac{1}{q}+1}\\ &= - \frac{\frac{pq+1}{pq}}{\frac{1-q+p-pq}{pq}}\\ &= \frac{pq+1}{pq-p+q-1} \end{align*}from which it is clear that $\frac{c-m}{t-h} = -\overline{\left(\frac{c-m}{t-h}\right)}$. Thus, $\frac{c-m}{t-h} \in i\mathbb{R}$ which proves that $CM \perp TH$ as desired.