Question

In ∆ABC, AD is the median through A and E is the mid-point of AD. BE produced meets AC in F. Prove that:

AF = 1/3 AC.​

Answer 1

$\huge\mathfrak\purple{Bonjour!!}$

$\huge\bold\pink{Solution:-}$

⛄ GIVEN:-

ABC is a ∆le. AD is the median through A and E is the mid-point of AD. BE is produced to F and it meets AC.

⛄ TO PROVE:-

AF = 1/3 AC.

⛄ CONSTRUCTION:-

Through D, draw DK || BF.

⛄ PROOF:-

In ∆ADK, E is the mid-point of AD and EF || DK.

Therefore,

F is the mid-point of AK.

=> AF = FK .....→ (i)

Now,

In ∆BCF, D is the mid-point of BC and DK || BF,

Therefore,

K is the mid-point of FC.

Therefore,

FK = KC ....→ (ii)

Now,

From (i) and (ii), we get,

AF = FK = KC .....→(iii)

Now,

AC = AF + FK + KC

=> AC = AF + AF + AF [Using (iii)]

=> AC = 3(AF)

=> AF = 1/3 AC

⛄ HENCE PROVED!!!

____________________________________

✔✔✔✔✔✔✔✔✔✔

Hope it helps....❣❣❣

⭐❤✨♥⭐❤✨♥⭐

Be Brainly....✌✌✌

♣ WALKER ♠

Answer 2

Answer:

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