Question

plz answer this question......

Answer 1

Answer:

Hi your answer is as follows

Step-by-step explanation:

Given:

AD is the median of ΔABC and E is the midpoint of AD

Through D

draw DG || BF

In ΔADG

E is the midpoint of AD and EF || DG

By converse of midpoint theorem we have

F is midpoint of AG and AF = FG  ..............(1)

Similarly, in ΔBCF

D is the midpoint of BC and DG || BF  

G is midpoint of CF and FG = GC ..............(2)

From equations 1 and 2

we will get

AF = FG = GC ........(3)

AF + FG + GC = AC

AF + AF + AF = AC (from equation  3)

3 AF = AC

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

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