Question

THE MEDIAN OF TRIANGLE ABC INTERSECT EACH OTHER AT POINT G.IF ONE OF ITS MEDIAN IS AD,PROVE THAT AREA OF TRIANGLE ABD IS EQAUL TO THRICE OF AREA OF TRIANGLE BGD

Answer 1

G is the centroid, so AG = 2GD

Answer 2

∆ABD = AD * BD sin(theta)
∆BGD = BD * GD sin(theta)
\[\frac{∆ABD}{∆BGD} = \frac{AD*BD*\sin \theta}{BD*GD*\sin \theta} = \frac{AD}{GD} = \frac{AG + GD}{GD} = \frac{3GD}{GD} = 3\]

(QED)