Question

in triangle ABC, AD is median through A and E is mid point of AD . BE is produced to meet AC in F. then prove that AF=1/3AC

Answer 1

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 ③

AF=AC

AF=(1/3)AC

~hope it helps

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Answer 2

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 will get

AF=FG=GC ─────── ③

AF+FG+GC=AC

AF+AF+AF=AC ────── from ③

AF=AC

AF=(1/3)A

Step-by-step explanation:

hope this helps you!!