Let $M$ be the tangency point of the circle with $AB$, we define the same way $N(N \in BC)$,$P(P \in AC)$,$Q(Q \in DE)$. Also $MB=BN=NC=CP=c$, $MD=DQ=a$, $QE=EP=b$, $AD=c-a$, $AE=c-b$. Then our statement is equivalent to show:$3ab=c^{2}-ac-bc$ By cosine law in triangle $AED$: $(b+c)^{2}=(c-a)^{2}+(c-b)^{2}-2cos60^\circ(c-a)(c-b)$, and the result follows.

With the last notations. $BE$ cut $CD$ at $J$ and $AJ$ cut $BC$ at $K$, then: $\frac{AD}{DB}+\frac{AE}{EC}=\frac{AJ}{JK}$ But $BE, CD, MP$ concur at $J$, now $J$ is the midpoint of $AK$,i.e. $AJ=JK$, therefore: $\frac{AD}{DB}+\frac{AE}{EC}=1$

Dear Mathlinkers, revisiting the last proof, we have 1. the relation of van Aubel 2. the Newton theorem 3. the midline. Sincerely Jean-Louis

This problem is a special case of the problem Modern Sangaku