Take $AD$ the third altitude, $D\in BC$. CX being tangent to that circle, $BX^2=BA\cdot BC_0\ (\ 1\ )$ ; similarly, $CX^2=CA\cdot CB_0\ (\ 2\ )$, but $BA\cdot BC_0=BC\cdot BD\ (\ 3\ )$ and $CA\cdot CB_0=BC\cdot CD\ (\ 4\ )$ hence $BX^2-CX^2=BC\cdot (BD-CD)= AB^2-AC^2$ or $AX\bot BC$. Best regards, sunken rock

$A{B^2} - X{B^2} = A{B^2} - AB \cdot B{C_0} = AB \cdot A{C_0}$，$A{C^2} - X{C^2} = A{C^2} - AC \cdot C{B_0} = AC \cdot A{B_0}$，Because $AB \cdot A{C_0} = AC \cdot A{B_0}$，so$A{B^2} - X{B^2} = A{C^2} - X{C^2}$, hence $AX \bot BC$.

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Dear yunxiu, I can't make the connection between $ A{B^{2}}-X{B^{2}}= A{C^{2}}-X{C^{2}} $ and $ AX\bot BC $.Is there a practicular formula for that? Best regards

At above: Suppose you have AB _|_ CD. Let their intersection point be P. Then you can use the Pythagorean Theorem to show that $AB^2 + CD^2 = AD^2 + BC^2$. If you start with the relation $AB^2 + CD^2 = AD^2 + BC^2$, then let the angle APB be $\theta$. You can use the Law of Cosines to backtrack and show that $\theta$ must be equal to $90^\circ$.

dragon96 wrote: $AB^2 + CD^2 = AD^2 + BC^2$. I think the above is true if and only if triangles APD and CPB are similar,isn't it?

No, consider this: If the two triangles are similar, here's a counter example: [asy][asy]defaultpen(linewidth(0.7)); pair A=origin, B=(12,0), C=(12,5), D=(0,5), P=(6,2.5); draw(A--B--C--D--A--C^^B--D); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$",D, NW); label("$P$", P, N);[/asy][/asy] If the relation holds, here's a counter example: [asy][asy]defaultpen(linewidth(0.7)); pair P=origin, B=(8,0), A=(0,3), C=(0,-14), D=(-13,0); draw(A--B--C--D--A--C^^B--D); label("$A$", A, N); label("$B$", B, E); label("$C$", C,S); label("$D$",D, W); label("$P$", P, NE, fontsize(10)); markscalefactor=0.1; draw(rightanglemark(A,P,D));[/asy][/asy] From the second diagram, it should hopefully clear why if they're perpendicular, the relation holds.

But look: $ AB^{2}+CD^{2}= AD^{2}+BC^{2} $ <=> $ AP^{2}+BP^{2}+2AP.BP+DP^{2}+CP^{2}+2BP.CP= AP^{2}+DP^{2}+BP^{2}+CP^{2} $ <=> $ AP.BP=CP.DP $ and with the right angle we get the two similar triangles.Where's the mistake?(if $ D $ and $ C $ are above $ P $ and $ D $ is above $ C $,which is the case in our problem)

Amir Hossein wrote: Let $ABC$ be an acute triangle. Denote by $B_0$ and $C_0$ the feet of the altitudes from vertices $B$ and $C$, respectively. Let $X$ be a point inside the triangle $ABC$ such that the line $BX$ is tangent to the circumcircle of the triangle $AXC_0$ and the line $CX$ is tangent to the circumcircle of the triangle $AXB_0$. Show that the line $AX$ is perpendicular to $BC$. $$BX^2=BC_0 \cdot BA$$$$CX^2=CB_0 \cdot CA$$$$AB^2-BX^2=AC^2-CX^2$$Carnot theorem.

$AA_0$ altitude. $X$ is intersection $AA_0$ and $(BCB_0C_0)$.

By the Perpendicularity Lemma, it suffices for $$XB^2-XC^2 = AB^2-AC^2 \iff BC_0 \cdot BA - CB_0 \cdot CA = AB^2-AC^2.$$But this is clearly true as $AC_0 \cdot AB = AB_0 \cdot AC$.