$\Delta CMP,\Delta CNQ$ are isosceles and, if $X, Y$ are midpoints of $MP, NQ\implies CX\bot AM, CY\bot BN$ and, since $AM, BN$ are angle bisectors, and if $D\in AB\cap CX, E\in AB\cap CX$, then $XY$ is midline of $\Delta CDE$, and we are done. Besat regards, sunken rock

It's just angle chase. Let $X,Y$ be midpoints of $MP$ and $NQ$ and $I$ be incenter of triangle $ABC$. Quadrilateral $CXIY$ is cyclic, so the result follows.

$X$-middle $PM$. $Y$-middle $QN$. We must prove $XY^{-}.CH^{-}=0$ $XY=\frac{QP^{-}+NM^{-}}{2}$ This way very easy.

an other way: let $X,Y$ be midpoint of the segments $PM,QN$, respectively. We have that $AHXC$ and $BHYC$ are cyclic. [because, $ \angle CXA= \angle CYB= \angle 90^\circ $ .] Hence, $XCYH$ is deltoid $ \Rightarrow $ $XY$ - perpendicular bisector of the segment $CH$ $ \Rightarrow $ $XY \| AB$.

I just bashed to calculate the lengths of $IX$ and $IY$ and then showed that $\frac{IX}{IY}=\frac{AI}{BI}$ from which the result follows.... Believe,me...its too easy for even to appear as the first problem in a TST...

Dear Mathlinkers, this problem can be also solved with the help of the Lascases theorem... See : http://perso.orange.fr/jl.ayme vol. 4 An unlikely concurrence. Sincerely Jean-Louis

This problem is really easy, I found 3 different solutions. I am giving one which is not posted. Well, let X and Y be on AB such that NX is perpendicular to AB and MY is perpendicular to AB. We clearly need to prove that NX+HQ=MY+HP. It is true because they are equal to CH, because CPYM and NXQC are parallelograms, which comes from few angle bisector theorems and Thales In triangles AHC and CHB.

sunken rock wrote: $\Delta CMP,\Delta CNQ$ are isosceles and, if $X, Y$ are midpoints of $MP, NQ\implies CX\bot AM, CY\bot BN$ and, since $AM, BN$ are angle bisectors, and if $D\in AB\cap CX, E\in AB\cap CX$, then $XY$ is midline of $\Delta CDE$, and we are done. Besat regards, sunken rock could you please justify what allowed you to say that it's the midline ?

omarius wrote: could you please justify what allowed you to say that it's the midline ? One can show that $\angle BDC=\angle BCD \implies BD=BC \implies DY=DC$, similarly $CX=XE$.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%207.pdf p. 67... Sincerely Jean-Louis

Let $U$, $V$ be the midpoint of $BC$, $AC$ and $S$, $T$ be the midpoint of $PM$, $QN$ respectively. It is sufficient to show $S\in UV$, because $UV\parallel AB$. Claim: The point $S$ lies on $UV$ i.e. on the $C-$midline. Proof. Notice that $\angle CMP=\angle CPM=90^\circ-\angle A/2$. So $\triangle CPM$ is isosceles. Therefore $CS\perp PM$ since $S$ is the midpoint of $PM$. So $AHSC$ is cyclic with center $V$. Hence $\angle VSA=\angle VAS=\angle SAH$, implies $VS\parallel AB$. That means $S$ lies on $UV$. $\square$ Similarly $T$ lies on $UV$. As desired.

$\color{red} \textbf{Geo Marabot Solve 2}$ $\triangle CMP$ and $\triangle CNQ$ are isosceles hence $CX \perp AM, CY \perp BN$ If $$CX \cap AB \equiv R, CY \cap AB \equiv S \implies XY \parallel RS \equiv AB \blacksquare$$

Coordinates! We employ cartesian coordinates. Set $C(0,0),A(0,a),B(1,0)$. Then by Angle-Bisector Theorem and Pythagoras theorem coordinates of points $M$ and $N$ are $\left(\dfrac{a}{a+\sqrt{a^{2}+1}},0\right)$ and $\left(0,\dfrac{a}{1+\sqrt{a^{2}+1}}\right)$ respectively. The equation of line $CD$ is $y=\dfrac{x}{a}$. The equation of line $AM$ is $y=-(a+\sqrt{a^{2}+1})x+a$. The equation of line $BN$ is $y=\left(\dfrac{-a}{1+\sqrt{a^{2}+1}}\right)x+\left(\dfrac{a}{1+\sqrt{a^{2}+1}}\right)$. Thus we get, $P \equiv \left(\dfrac{a^{2}}{\sqrt{1+a^{2}}(a+\sqrt{1+a^{2}})},\dfrac{a}{\sqrt{1+a^{2}}(a+\sqrt{1+a^{2}})}\right)$ and $Q \equiv \left(\dfrac{a^{2}}{\sqrt{1+a^{2}}(1+\sqrt{1+a^{2}})},\dfrac{a}{\sqrt{1+a^{2}}(1+\sqrt{1+a^{2}})}\right)$. Now let midpoints of $QN$ and $PM$ be denoted by $K$ and $L$ respectively, then: $K \equiv \left(\dfrac{a^{2}}{2\sqrt{1+a^{2}}(a+\sqrt{1+a^{2}})},\dfrac{a}{2\sqrt{1+a^{2}}}\right)$ and $L \equiv \left(\dfrac{a}{2\sqrt{1+a^{2}}},\dfrac{a}{2\sqrt{1+a^{2}}(a+\sqrt{1+a^{2}})}\right)$. Thus slope of the line passing through $K$ and $L$ is: $m_{KL}=\dfrac{\dfrac{a}{2\sqrt{1+a^{2}}}\left(1-\dfrac{1}{a+\sqrt{1+a^{2}}}\right)}{\dfrac{a}{2\sqrt{1+a^{2}}}\left(-1+\dfrac{a}{1+\sqrt{1+a^{2}}}\right)}$ $\implies m_{KL}=\dfrac{1-(\sqrt{1+a^{2}}-a)}{\dfrac{\sqrt{1+a^{2}}-1}{a}-1}=-a$. Also $m_{AB}=\dfrac{a-0}{0-1}=-a$. As the slopes of these lines are equal hence they are parallel. (QED) $\blacksquare$