Name: Circumcentre of a Triangle
Uploaded: 2019-07-19
Duration: 7 min 3 s
Description: The circumcentre of a triangle is one of the important topics in Geometry. Learn about this concept along with examples by watching this video.

Question

Prove that the distance of the circum-centre from any side of a triangle is half of its distance of its orthocentre from the vertex just opposite to that side of the triangle.

Accepted Answer

Let $O$ be the circumvented of triangle $A B C$ and define $O H$ as the vector sum $O A + O B + O C$ .

Then $AH = OH - OA = OB + OC = 2 OM$ , where $M$ is the midpoint of $B C$ .

Now as $O B$ and $O C$ are radii of the circumscribed circle, their lengths are equal; so the triangle $O B C$ is isosceles and $O M$ , hence also $A H$ , is perpendicular to $B C$ .

So $H$ lies on the height from $A$ .
But by the same reasoning, $H$ also lies on the other two heights.

This proves that the three heights meet at a common point, called the orthocenter, and that its distance from $A$ , hence from any vertex, is twice the distance from O to the opposite side. (It also proves that the barycentre $G$ , defined by $O G = \frac{(O A + O B + O C)}{3}$ for any point $O$ , lies on $O H$ and divides it on a $1 : 2$ ratio.)

Therefore, the distance of the circum-centre from any side of a triangle is half of its distance of its orthocenter from the vertex just opposite to that side of the triangle.

Prove that the distance of the circum-centre from any side of a triangle is half of its distance of its orthocentre from the vertex just opposite to that side of the triangle.

Important Questions on Theorems on Concurrence

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