Let $ABC$ be a triangle such that $2AB=AC+BC$. Show that the incentre, the circumcircle, midpoint of $AC$ and midpoint of $BC$ lie on a circle.
I reduced the question to prove that both midpoints, incentre and vertex $C$ lie on a circle. So we have to prove that $\angle{MIN}=180-\angle{C}$. But I am not getting this.
Best Answer
Let the midpoints of $CB$ and $CA$ be $A'$ and $B'$. Now we place a point on $D$ on $AB$ such that $AD=AB'$. And as $2c=a+b$, we also have $BD=BA'$.
Thus, by SAS congruence, $\triangle BIA'\cong \triangle BID$ and $\triangle AIB'\cong \triangle AID$. Thus, $\angle BDI= \angle IA'B$ and $\angle ADI =\angle IB'A$. Hence, $\angle IB'A=180^{\circ}-\angle IA'B=\angle IA'C$.
Thus, $C$, $B'$, $A'$ and $I$ lie on a circle.