# incircle tanget to triangle at D and incirles of ADC ADB

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

• Could You put the original wordings here and is there any picture? – Rezha Adrian Tanuharja Mar 22 '20 at 9:24

## 2 Answers

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

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