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Let's say we have isosceles triangle $ABC$ s.t. $AC=BC$. Let's define the middle of $AB$ as point $M$. On $AB$ we choose a random point $X$. We inscribe a circle with center $O_1$ in triangle $AXC$ and another circle with center $O_2$ in triangle $XBC$. Proof that the angle $O_1MO_2$ is exactly $90°$.

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  • $\begingroup$ Which sides of the trianlge are equal? $\endgroup$ – Andreas Ch. Feb 24 '17 at 13:45
  • $\begingroup$ Sorry for not clarifying that. I have updated the question (AC=BC). $\endgroup$ – orim Feb 24 '17 at 13:47
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Let $c_1$ be the incircle in triangles $AXC$, whose incenter is $O_1$, and let $c_2$ be the incircle in triangles $BXC$, whose incenter is $O_2$. Observe that line $CX$ is the internal common tangent to the circles $c_1$ and $c_2$, while $AB$ is the external common tangent to $c_1$ and $c_2$. Let line $p$ be the other common internal tangent of circles $c_1$ and $c_2$ and let $p$ intersect $AB$ at point $N$.

I will show that $N$ is the midpoint of $AB$ and thus $N \equiv M$.

First, some notation: Let $R_1$ be the point of tangency of edge $AC$ and circle $c_1$, let $S_1$ be the point of tangency of segment $CX$ and circle $c_1$, and let $T_1$ be the point of tangency of edge $AB$ and circle $c_1$. Analogously, let $R_2$ be the point of tangency of edge $BC$ and circle $c_2$, let $S_2$ be the point of tangency of segment $CX$ and circle $c_2$, and let $T_2$ be the point of tangency of edge $AB$ and circle $c_2$. Furthermore, let the line $p$ touch circle $c_1$ at point $S_1'$ and circle $c_2$ at point $S_2'$

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Without loss of generality, assume $AX > BX$.

Let $AC = BC = l$ and let $S_1S_2 = s$. By tangency, $CR_1 = CS_1 = h$ and $S_1'S_2' = S_1S_2 = s$. Hence $CR_2 = CS_2 = CS_1+S_1S_2 = h+s$. Therefore $AR_1 = AC - CR_1 = l-h$ and $BR_2 = BC - CR_2 = l - h - s$. Again by tangency $$AT_1 = AR_1 = l-h \,\,\, \text{ and } \,\,\, BT_2 = BR_2 = l-h-s$$ Set $NT_1 = t$. Then by tangency, $NS_1' = NT_1 = t$ and $NT_2 = NS_2'$. However, $NS_2' = NS_1' + S_1'S_2' = t+s$ so $NT_2 = NS_2'=t+s$.

Now let us add the lengths of the appropriate segments $$AN = AT_1 + NT_1 = l-h + t$$ $$BN = BT_2 + NT_2 = l-h-s + t+s = l-h+t$$ Thus, we can conclude that $AN = l-h+t = BN$ and so $N$ is the midpoint of segment $AB$, which means that $N \equiv M$.

To conclude the proof, observe that $$\angle \, O_1MS_1' = \angle \, O_1MT_1 = \frac{1}{2} \, \angle \, T_1MS_1'$$ $$\angle \, O_2MS_2' = \angle \, O_2MT_2 = \frac{1}{2} \, \angle \, T_2MS_2'$$ Since the points $T_1, M$ and $T_2$ lie on the segment $AB$, $$ \angle \, T_1MS_1' + \angle \, T_2MS_2'= \angle \, T_1MT_2 = 180^{\circ}$$ so $$\angle \, O_1MO_2 = \angle \, O_1MS_1' + \angle \, O_2MS_2' = \frac{1}{2} \, \angle \, T_1MS_1' + \frac{1}{2} \, \angle \, T_1MS_1' = $$ $$= \frac{1}{2} \, \big(T_1MS_1' + \angle \, T_1MS_1' \big) = \frac{1}{2} \, 180^{\circ} = 90^{\circ}$$

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  • $\begingroup$ Thank you for your solution! I solved the problem using Cartesian coordinate system and the calculations were overheat. I was looking for some other approach (like yours) which will expose the hidden structure, making this angle right. $\endgroup$ – orim Feb 25 '17 at 5:42
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We may recall that if $ABC$ is a triangle with side lengths $a,b,c$, its incenter is given by $\frac{aA+bB+cC}{a+b+c}$. We may assume without loss of generality that $M=(0,0),B=(1,0),A=(-1,0),$ $C=(0,h),X=(x,0)$, then compute $$ O_1=\frac{X\sqrt{1+h^2}+B\sqrt{x^2+h^2}+C(1-x)^2}{\ldots},\qquad O_2=\frac{X\sqrt{1+h^2}+A\sqrt{x^2+h^2}+C(1+x)^2}{\ldots} $$ and easily check that $O_1 M \perp O_2 M$ as wanted.

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  • $\begingroup$ Thank you for your solution! I solved this problem using exactly the same approach (unfortunately, I didn't chosen $M$ as center of the coordinate system and the calculations were overheat).I was wondering - is there some other, more elegant way to solve this relation. A little correction: the coefficients in front of C are $1-x$ and $1+x$. The system is not allowing me to mark your solution as correct answer, too. $\endgroup$ – orim Feb 25 '17 at 5:39

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