Proof that the equilateral triangle incentre and the circumcentre are the same point

**MA2012** · August 30th 2012, 11:58 PM

Hello

I have a question on my maths assignment I need some help with.

The question says - Make a conjecture about the nature of a triangle where the Incentre and the Circumcentre are the same point. Prove your conjecture.

I know that its going to be an equilateral triangle, and I think I have to prove it by contradiction.

thanks

**MaxJasper** · August 31st 2012, 12:19 AM

Incenter and circumcenter to coincide means that from the same point inside a triangle you can draw a circle that passes through each vertex and another smaller circle that is inscribed inside the triangle...draw these and prove that right angles generated by radius of inscribed circle on the sides of triangle cannot be right angles!

**kalyanram** · September 13th 2012, 12:26 AM

Proof that the equilateral triangle incentre and the circumcentre are the same point-incir.png

Given $O$ is the Incenter and the Circumcenter of the $\Delta ABC$ . Let $D,E,F$ be the foot of the perpendiculars from $O$ on sides $AB$ , $BC$ , $CA$ respectively. And $AD', BE', CF'$ be the angular bisectors of $\angle A$ , $\angle B$ , $\angle C$ respectively.

In $\Delta AOB$ , $OA = OB =R$ (circumradius), $\therefore \Delta OAB$ is isosceles $\therefore\angle OAB = \angle OBA \implies \frac{1}{2} \angle A = \frac{1}{2} \angle B$ (as $OA$ and $OB$ are angular bisectors also) $\imples \angle A = \angle B$
Similarly in $\Delta AOC$ we get $\angle C = \angle B$ .

$\therefore \Delta ABC$ is equilateral triangle.

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