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$ABC$ is a triangle the circle is tangent to $BC$ at $D$ prove that the incircles of $\triangle ABD$ and $\triangle ACD$ are tangent to each other.

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What i tried is calling the smaller incircles tangent to $AC$ and $AB$ at $E$ and $F$, respectively. Then i need to prove $AE=AF$.

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  • $\begingroup$ Could You put the original wordings here and is there any picture? $\endgroup$ – Rezha Adrian Tanuharja Mar 22 '20 at 9:24
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Let touch-points of the incircles of $\Delta ACD$ and $\Delta ADB$ to $AD$ be $P$ and $Q$ respectively.

Thus, in the standard notation we obtain: $$DP=\frac{AD+CD-AC}{2}=\frac{AD+\frac{a+b-c}{2}-b}{2}=\frac{AD+\frac{a+c-b}{2}-c}{2}=DQ$$ and we are done!

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Repeatedly use that the distances from any point to the two tangent points of a circle is equal,

$$\begin{array} AAE - AF & = (AY+YE)-(AX+XF) \\ & =YE - XF\\ & = (CY - CE) - (BX - BF) \\ & = (CD - CP) - (BD - BQ) \\ & = DP - DQ \\ & = DT - DT = 0 \\ \end{array}$$

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