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Let $ABC$ be a triangle and let $D$ be a point on side $BC$. Show that the incircles of triangles $ABD$ and $ACD$ touch each other if and only if $D$ is the point of contact of the incircle of triangle $ABC$ with $BC$.

I tried some trigonometry but no success. Sides chasing didn't help either. Thanks.

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It is trivial that the incircles of $\triangle ABD$ and $\triangle ACD$ touch if and only if they hit $AD$ at the same point.

Let the incircle of $\triangle ABD$ hit $AD$ at $X_1$ and the incircle of $\triangle ACD$ hit $AD$ at $X_2$.

Therefore, $$\text{Two incircles touch each other} \iff X_1 = X_2 \iff AX_1=AX_2 \iff \frac{AB+AD-BD}{2}=\frac{AC+AD-CD}{2} \iff AB-AC=BD-CD \iff BD=s-b, CD=s-c \iff \text{D is the contact point of the incircle}$$

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  • $\begingroup$ Sorry, but what is s ? $\endgroup$ – Nizar Dec 1 '15 at 11:02
  • $\begingroup$ $s=\frac{1}{2}(a+b+c)$ $\endgroup$ – Gyumin Roh Dec 1 '15 at 11:05

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