2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ You must have meant circumcenter, I hope? $\endgroup$
    – Sawarnik
    Jan 17 '15 at 8:45
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.