Question

In triangle ABC , AD is a median and M is mid point of AD . When produced B meets AC at N . Prove that AN = ⅓ AC .

Answer 1

hey sis here is 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 will get
AF = FG = GC ........3
 AF + FG + GC = AC
AF + AF + AF = AC (from eu 3)
3 AF = AC
AF = (1/3) AC

hope it helps you dear

Answer 2

Given:- In triangle ABC
i) AD is Median
ii) M is the mid point of ADMy aMt No problem .

To prove that :- AN = AC/3

Construction :- Draw DK || BF

Proof:-
In triangle BNC,
D is the mid- point of BC.
and DK || BN
Hence, K will be the mid point of NC.
i.e. NK = KC ------ (1)

Now,
In triangle ADK,
M is the mid- point of AD
& MN || DK (By Construction)
Hence, N will be bisector of AK.
i.e. AN = NK --------(2)

From equation (1) & (2) We get,

=> AN= NK = KC

But,

=> AN+NK+KC = AC
=> AN+AN+AN = AC [ Because AN=NK=KC]
=> 3 AN = AC
=> AN = AC/3 or 1/3 AC

Hence Proved.

Be Brainly