# Orthocenter and midpoint proof

Let $H$ be the orthocenter of a triangle $\Delta ABC$. Let $M$ be the midpoint of $BC$, and let $E$, $F$ be the feet of the $B$ and the $C$ altitudes onto the opposite sides. Let $X$ be the intersection of ray $MA$ with the circumcircle of $BHC$.

Prove that

(1) $HX$, $EF$ and $BC$ concur at a point, say $P$;

(2) line $MX$ is perpendicular to the line $XH$, where $O$ is the center of the circumcircle; and

(3) also show that the line joining that point $P$ and $A$ is perpendicular to the line $HM$.

• Welcome to math.SE! You are more likely to get useful help on this site if you explain what you've tried so far, where you are stuck. Otherwise you are likely to get downvoted or have your question closed. – Tengu Dec 22 '17 at 4:09
• What and where is O? – Mick Dec 22 '17 at 4:53
• @Mick I guess $O$ is the center of the circle – user122049 Dec 22 '17 at 8:50
• Yes, O is the center of the circumcircle of the triangle BHC. – user061703 Dec 22 '17 at 9:51
• (2) is not true. The correct version is to prove that "Line MX is perpendicular to the line XH".. – Mick Dec 25 '17 at 5:07

Let $AD$ be the third altitude of $\triangle ABC$. Then, $DEF$ forms the pedal triangle which has the properties of having $H$ being its in-center. Therefore, $\theta = \theta_1$, for example. {Note: This part may not be necessary but I will leave it on for reference purpose.}

Extend $AD$ to cut the circle $BHC$ at $S$. Join $BS$ and $CS$. Then, $\alpha = \alpha_1 = \alpha_2$. This makes $BACS$ a kite. Therefore, $\omega = \omega_1$.

Construct $BT$//$AC$ cutting the circle $BHC$ at $T$. Join $CT$. All dark green shaded angles are equal. In addition, eventually, $\omega_1 = \omega_2$ (through subtended arcs). This means $BCTS$ is an isosceles trapezium with $BC$//$ST$. This proves $ABTC$ is a parallelogram with $AMT$ is a diagonal (i.e. $AMT$ is in fact a straight line.)

$DHXM$ is then cyclic because the purple shaded angles are all equal. This completes the proof of (2).

Then, $H$ is the orthocenter of $\triangle APM$. This completes the proof of (3).

For (1), we can argue in the following way:

a) Let $FE$ cut $AM$ at $U$. b) Draw $UV \bot PM$ and let it cut $PX$ at $W$.

So far we have fixed (i) two sides ($MP$ and $MU$); (ii) three points ($U$, $M$ and $P$); (iii) two possible altitudes ($UV$ and $PX$); and (iv) the possible orthocenter ($W$).

It is sufficient to claim (I) $PU$ is the third side of the triangle $MPU$; (II) For $\triangle MPU$, $PU$ and $PM$ are two of the adjacent legs with $PX$ being the included altitude; and (III) The three lines are concurrent at $P$.