Problem

Source:

Tags: geometry



In $\triangle ABC$, points $D$, $E$, and $F$ lie on sides $BC$, $CA$, and $AB$, respectively, such that each of the quadrilaterals $AFDE$, $BDEF$, and $CEFD$ has an incircle. Prove that the inradius of $\triangle ABC$ is twice the inradius of $\triangle DEF$.