The point of intersection of the common chords of three circles described on the three sides of a triangle as diameter is

Options

A. centroid of the triangle
B. orthocentre of the triangle
C. circumcentre of the triangle
D. incentre of the triangle

Answer

B

Explanation

Let ABC be a triangle and let AD, BE and CF be its altitudes. Then, `angleADB=pi//2` and `angleAEB=pi//2`. Therefore, the circle on AB and AC as diameters passes through D. Consequently, AD is the common chord of the circles on AB and AC as diameters.
Similarly, the other two altitudes BE and CF are the other two common chords. Thus, the point of intersection of the common chords is the point of intersection of AD, BE and CF i.e. the orthocentre of `DeltaABC`.