Inscribed circle cuts triange into 3 pairs of congruent segments?

How do you prove that if you inscribe a circle in a triangle $ABC$, and if $X_{AB}$, $Y_{BC}$, and $Z_{CA}$ are the points where the circle is tangent to the (sub-scripted) side of the triangle, then $AX_{AB}$ is congruent to $AZ_{CA}$, $BX_{AB}$ is congruent to $BY_{BC}$, and $CY_{BC}$ is congruent to $CZ_{CA}$?

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Let $O$ be the centre of the circle,

Consider the $\Delta OAZ_{CA}$ and $\Delta OAX_{AB}$

We have $OX_{AB}=OZ_{CA}$=radius of the circle, angle$(OZ_{CA}A)=$angle$(OX_{AB}A)=\pi/2$(from tangency) and $OA$ is same in both the triangles . So we have

$\Delta OAZ_{CA} \simeq \Delta OAX_{AB}$

And from this it follows ,$AZ_{CA}=AX_{AB}$,

Similar arguement proves the other equalities.

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