Question

answer it please fast​

Answer 1

Answer:

Step-by-step explanation:

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

Hope it helps u mark as brainliest please

Answer 2

           

As you have drawn in the figure, draw line DQ, parallel to BF.

Basic proportionality theorem states that, a line, parallel to a side of a triangle, cuts the other two sides in the same ratio.

Consider ΔBFC.

As BF ║ DQ,

FQ : QC = BD : DC = 1 : 1     →     (1)          [∵ BD = DC]

Consider ΔAQD.

As BF ║ DQ,

AF : FQ = AG : GD = 1 : 1     →     (2)          [∵ AG = GD]

From (1) and (2),

FQ : QC = AF : FQ = 1 : 1

∴ AF = FQ = QC

AF + FQ + QC = AC

⇒ AF + AF + AF = AC

⇒ 3 AF = AC

⇒ AF = 1/3  AC

Hence proved!!!

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