In an equilateral triangle, the circumcenter coincides with which of the following?
The centroid of the triangle
Consider any single vertex of an equilateral triangle. From it, the median (to the opposite midpoint), the altitude (perpendicular to the opposite side), the internal angle bisector, and the perpendicular bisector of the opposite side are all the very same line.
This happens because the triangle has full three-fold symmetry: opposite each vertex the side is symmetric, so all these special lines collapse together.
The circumcentre is the common point of the three perpendicular bisectors, and the centroid is the common point of the three medians.
Since perpendicular bisectors and medians coincide line-for-line, their intersection points must coincide too.
In fact all four classical centres, the circumcentre, incentre, centroid and orthocentre, land on this one point in an equilateral triangle.
So the circumcentre is located exactly at the centroid, at a distance \(\tfrac{2}{3}\) of each median from the vertex.
Hence the circumcenter coincides with the centroid of the triangle.
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Important Questions from Triangles, Congruence and Similarity
Angle between the internal bisectors of two angles ∠B and ∠C of a ΔABC is 132°, then the value of ∠A is
In ΔPQR, PQ = PR and S is a point on QR such that ∠PSQ = 96° + ∠QPS and ∠QPR = 132°. What is the measure of ∠PSR?
In Δ ABC, ∠A = 50°. If the bisectors of the angle B and angle C, meet at a point O, then ∠BOC is equal to:
Triangle ABC is right angled at B. BD is an altitude intersecting AC at D. If AC = 9 cm and CD = 3 cm. then find the measure of AB (in cm).
The base and altitude of an isosceles triangle are 10 cm and 12 cm respectively. Then the length of each equal side is:
