Question

Show that in an equilateral triangle, circumcentre, orthocentre and incentre overlap each other.

Answer 1

CIRCUMCENTRE: Point of Concurrence of perpendicular bisector.

ORTHOCENTRE: Point of Concurrence of altitude.

INCENTRE: Point of Concurrence of angle bisector.

Let $AD$ be altitude.

In $△$ ABD

$\therefore \angle BOA={90}^0$

$\therefore \angle BOA=180-(20+60)={30}^0$

Similarly in $△ADC$,

$\angle BAC={30}^0$

$\therefore AD$ is an angle bisector.

In $△ABD$

$\sin 30=\frac{BD}{BA}$

In $△ADC$

$\sin 30=\frac{DC}{AC}$

$\therefore BD=DC$

$\therefore$ $D$ is the midpoint of $BC$

$\therefore AO$ is perpendicular bisector.

So, if $AD$ is an altitude,

$i)AD$ is an angle bisector.

$ii)AO$ is perpendicular bisector.

$\therefore$Circumcenter, incentre and orthocenter overlap each other.