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The point of intersection of the common chords of three circles described on the three sides of a triangle as diameter isA. centroid of the triangleB. orthocentre of the triangleC. circumcentre of the triangleD. incentre of the triangle |
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Answer» Correct Answer - B 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`. |
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