Question 6 how to solve

Dear Student,

Given that:  AD is the median and E is the mid-point of AD in triangle ABC.

To Prove:  $\mathrm{AF} = \frac{1}{3}\mathrm{AC}$

Construction:    Draw DG parallel to EF.

Proof:

Basic Proportionality Theorem (BPT): If a line is drawn parallel to one side of the triangle to intersect the other two sides at distinct points. Then the other two sides are divided in the same ratio.

In,

Since, D is the mid- point of BC, so, BD = CD

Thus,

FG = CG   …… (1)

In,

Since, E is the mid-point of AD, so, AE = ED

Thus,

AF = FG  ……. (2)

From equation (1) and (2), we have,

AF = FG = CG

$\Rightarrow \mathrm{AF} = \frac{1}{3}\mathrm{AC}$

Regards