Question

In the figure, D is the midpoint of side BC of triangle ABC. G is the midpoint of seg AD. seg BG when produced meets side AC at F. Prove AF =1/3AC using BPT theorem.​

Answer 1

Given AD is the median of ΔABC.
E is the midpoint of AD, also BG meets AD at F after producing BG
Draw DE||BF
In ΔADG,
G is the midpoint of AD and GF||DE.

By converse of midpoint theorem,

we have
F is the midpoint of AE
That is AF = FE --- (1)

Similarly,

in ΔBCF,

D is the midpoint of BC and DE||BF

Therefore,

E is the midpoint of FC

Hence

FE= EC ---(2)

From equations (1) and (2), we get
AF = FE = EC --- (3)

we have,


AF + FG + GC = AC

⇒ AF + AF + AF = AC [From (3)]


3AF = AC



Hence AF = (1/3) AC

Answer 2

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