Question

SOLVE THE PROBLEM.
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Answer 1

☀ SOLUTION:-☀

$\implies$ let G be the midpoint of FC and joint DG.

$\implies$in ΔBCF,

$\implies$ G is the mid point of FC and D is the mid point of BC.

$\implies$ thus DG || BF , DG || EF

$\implies$ now, in ΔADG,

$\implies$ E is the mid point of AD and EF is parallel to DG.

$\implies$ F, is the mid point of AG.

$\implies$ AF=FG=GC (G is the mid point of FC.)

$\implies$ $\large\red{\tt{\boxed{\boxed{hence, AF =1/3AC. }}}}$

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$\large\blue{\mathbf{\underline{\underline{SO,\: Final \:answer \: is \: option 'B'. }}}}$

Answer 2

ॐ $\huge{\orange{\underline{\pink{\mathscr{Radhe}}}}}$$\huge{\orange{\underline{\green{\mathscr{Radhe}}}}}$ॐ

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⭐ $\huge{\orange{\underline{\red{\mathbb{Solution}}}}}$⭐

$\implies$ Let the G midpoint of the FC and joint DG.

$\implies$ In the ∆BCF, G is the midpoint of FC & D is the midpoint of BC.

$\implies$ .°. DC || BF, DG || EF.

$\implies$ Here, in ∆ADG,

$\implies$ E is the mid point of AD & EF || DG

$\implies$ F, is the midpoint of AG

$\implies$ AF = FG = GC

$\implies$ ( G is the midpoint of FC )

$\huge{\fbox{\fbox{\bigstar{\red{AF = 1/3 AC}}}}}}$

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$\huge{\orange{\underline{\pink{\mathscr{(b)\: option\: is\:your}}}}}$

$\huge{\orange{\underline{\red{\mathbf{final\:Answer}}}}}$