Question 62

D is the midpointof side BC of $$\triangle ABC$$. Point E lies on AC such that $$CE = \frac{1}{3}AC$$. BE and AD intersect at G. What is $$\frac{AG}{GD}$$?

Solution

D is mid point of BC.

To apply the mid poingt theorem in ADM,

$$\frac{AG}{GD} = \frac{AE}{EM}$$

AE = $$\frac{2AC}{3}$$

EC = $$\frac{AC}{3}$$

EM = $$\frac{EC}{2}$$ = $$\frac{\frac{AC}{3}}{2}$$

= $$\frac{AC}{6}$$

$$\frac{AG}{GD} = \dfrac{\frac{2AC}{3}}{\frac{AC}{6}}$$ = 4 : 1

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