Question

in a triangle ABC it is a median and he is a midpoint of median ad a line through B and C meet AC at point f prove that AC is equal to 3 AF​

Answer 1

Answer:

Step-by-step explanation:

AD is the median of triangle ABC.

E is the mid point of AD.BE produced meet ad at F.

RTP ::AF=1/3AC.

Construction:From point D draw DG parallel to BF.

solution : 

By mid point theorem we got  F as the mid point of AG

-> AF=AG --------1

 G as the mid point of CF

-> FG=GC--------------2

by one and two

AF = FG= GC  ---------- 3

now ,   AF + FG + GC =AC

by the 3 rd equation

AF+AF+AF=AC

3×AF =AC

 so  1/3 ac = af

So AC=3AF.

hope this helps you

Answer 2

SOLUTION

Given,

AD is the median of ∆ABC. E is the midpoint of AD

BE produced meets AD at F

To prove:

AF= 1/3AC

Construction:

From point D, draw DG||BF.

Proof:

In ∆ADG,E is the midpoint of AD&EF||DG

F is the midpoint of AG[converse of the midpoint theorem]

=) AF= FG...............(1)

In ∆BCF, D is the midpoint of BC&DG||BF

G is the midpoint of CF

=) FG= GC...............(2)

From (1) & (2), we get,

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

Now,

AF+ FG+GC= AC

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

=) 3AF= AC

=) AF= 1/3AC [Proved]

Hope it helps ☺️