1 point: the intersection point of the bisector of an angle is the center of the inscribed circle of a triangle-at the center, the distance from each side is equal.

2 points: the intersection of the vertical lines in the middle of the side is the center of the circumscribed circle of the triangle-the outer center, and its distance to each vertex is equal?

3 points: the intersection of the connecting line of a triangle and the midpoint of the corresponding side is the center of gravity of the triangle-its distance from the vertex is twice as long as its distance from the midpoint of the opposite side?

4 points: the intersection of three high lines of a triangle is called the vertical center.

This problem involves two points, that is, point O is the outer center of △ABC and point H is the center of △ABC.

As shown in the figure, connect OA, OB, OC and CH.

Vector OH= vector OA+ vector AH? ( 1)

Vector OH= vector OB+ vector BH? (2)

Vector OH= vector OC+ vector CH(3)

Add three types and you get.

3*? Vector OH= (vector OA+ vector OB+ vector OC)+ (vector AH+ vector BH+ vector CH)

Because the vector sum from the vertical center to the vertex of the triangle is 0, that is? Vector AH+ vector BH+ vector CH=0

So, the vector OH= 1/3? * (vector OA+ vector OB+ vector OC)