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It is really hard to write Tex on the cellphone! Claim: Let $B'$ and $C'$ be the diametrical points of $B$ and $C$.Then we have $C',E,H$ and $B',F,H$ collinear. Proof: Let $C'H \cap AB=E'$ and $B'H \cap AC=F'$.We are going to prove that in fact $B,D,F'$ collinear,then it is easy to get that $F=F'$ and $E=E'$.Notice that $$\frac{B'B}{BD}\cdot \frac{DO}{DH}\cdot \frac{F'H}{F'B'}=2\cdot \frac{OM}{AH}\cdot \frac{AH}{B'C}=1$$. Then by Menelaus's Theorem we have $B,D,F'$ collinear. Back to the problem.We have $\angle BYH=\angle CXH=\frac{\pi}{2}$.Let $BY\cap CX=Z$ ,then $H,X,Y,Z$ is on a circle while $HZ$ its diameter.Now we just need to prove that $BC$ is the tangent line to the circle,then we are done.Just notice that $$\angle HYX=\overset{\frown}{B'C}+\overset{\frown}{CX}=\angle CHX$$,which means $A,H,Z$ collinear.$\blacksquare$

China1_mod_2 wrote: It is really hard to write Tex on the cellphone! Claim: Let $B'$ and $C'$ be the diametrical points of $B$ and $C$.Then we have $C',E,H$ and $B',F,H$ collinear. Proof: Let $C'H \cap AB=E'$ and $B'H \cap AC=F'$.We are going to prove that in fact $B,D,F'$ collinear,then it is easy to get that $F=F'$ and $E=E'$.Notice that $$\frac{B'B}{BD}\cdot \frac{DO}{DH}\cdot \frac{F'H}{F'B'}=2\cdot \frac{OM}{AH}\cdot \frac{AH}{B'C}=1$$. Then by Menelaus's Theorem we have $B,D,F'$ collinear. Back to the problem.We have $\angle BYH=\angle CXH=\frac{\pi}{2}$.Let $BY\cap CX=Z$ ,then $H,X,Y,Z$ is on a circle while $HZ$ its diameter.Now we just need to prove that $BC$ is the tangent line to the circle,then we are done.Just notice that $$\angle HYX=\overset{\frown}{B'C}+\overset{\frown}{CX}=\angle CHX$$,which means $A,H,Z$ collinear.$\blacksquare$ Can you explain on what triangle you used Menelaus'?

PNT wrote: Can you explain on what triangle you used Menelaus'? I think it's $\triangle OHB'$.

PNT wrote: China1_mod_2 wrote: It is really hard to write Tex on the cellphone! Claim: Let $B'$ and $C'$ be the diametrical points of $B$ and $C$.Then we have $C',E,H$ and $B',F,H$ collinear. Proof: Let $C'H \cap AB=E'$ and $B'H \cap AC=F'$.We are going to prove that in fact $B,D,F'$ collinear,then it is easy to get that $F=F'$ and $E=E'$.Notice that $$\frac{B'B}{BD}\cdot \frac{DO}{DH}\cdot \frac{F'H}{F'B'}=2\cdot \frac{OM}{AH}\cdot \frac{AH}{B'C}=1$$. Then by Menelaus's Theorem we have $B,D,F'$ collinear. Back to the problem.We have $\angle BYH=\angle CXH=\frac{\pi}{2}$.Let $BY\cap CX=Z$ ,then $H,X,Y,Z$ is on a circle while $HZ$ its diameter.Now we just need to prove that $BC$ is the tangent line to the circle,then we are done.Just notice that $$\angle HYX=\overset{\frown}{B'C}+\overset{\frown}{CX}=\angle CHX$$,which means $A,H,Z$ collinear.$\blacksquare$ Can you explain on what triangle you used Menelaus'? Just as @PhilippineMonkey says, we use Menelaus's Theorem on $\triangle OHB'$ and the line $BDF$.

P.S.

The problem is solved with Kaiwei Shao , one of my classmates who does excellent in geometry.

China1_mod_2 wrote: PNT wrote: China1_mod_2 wrote: It is really hard to write Tex on the cellphone! Claim: Let $B'$ and $C'$ be the diametrical points of $B$ and $C$.Then we have $C',E,H$ and $B',F,H$ collinear. Proof: Let $C'H \cap AB=E'$ and $B'H \cap AC=F'$.We are going to prove that in fact $B,D,F'$ collinear,then it is easy to get that $F=F'$ and $E=E'$.Notice that $$\frac{B'B}{BD}\cdot \frac{DO}{DH}\cdot \frac{F'H}{F'B'}=2\cdot \frac{OM}{AH}\cdot \frac{AH}{B'C}=1$$. Then by Menelaus's Theorem we have $B,D,F'$ collinear. Back to the problem.We have $\angle BYH=\angle CXH=\frac{\pi}{2}$.Let $BY\cap CX=Z$ ,then $H,X,Y,Z$ is on a circle while $HZ$ its diameter.Now we just need to prove that $BC$ is the tangent line to the circle,then we are done.Just notice that $$\angle HYX=\overset{\frown}{B'C}+\overset{\frown}{CX}=\angle CHX$$,which means $A,H,Z$ collinear.$\blacksquare$ Can you explain on what triangle you used Menelaus'? Just as @PhilippineMonkey says, we use Menelaus's Theorem on $\triangle OHB'$ and the line $BDF$.

P.S.

The problem is solved with Kaiwei Shao , one of my classmates who does excellent in geometry. Thank you so much for letting us know his name.

Let $L$ be the point on the circumcircle opposite to $B$. Using Desargues on $\triangle AMC$ and $\triangle HOL$, we get that $H,F,L$ collinear, hence $\angle HYB=\angle CXH=90^\circ$. Radical Axis on $(ABC),(HYB),(CHX)$ finishes.