Question

100 points please answer

Answer 1

hey friend!

here's your answer:

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 get

AF = FG = GC -------- (3)

now, we have, AF + FG + GC = AC

AF + AF + AF = AC [from (3)]

3 AF = AC

AF = (1/3) AC

i hope this will helps you

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Answer 2

here's your answer:

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 get

AF = FG = GC -------- (3)

now, we have, AF + FG + GC = AC

AF + AF + AF = AC [from (3)]

3 AF = AC

AF = (1/3) AC

thanks for asking be brainly           jaydeep