Question

Solution please it's urgent

Answer 1

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• For triangle ∆ABC ,AD is the median and E is the mid point of AD and BE is produced to F..prove that AF=(1/3)AC

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$\:\: \underline{\underline{\bf{\large\mathfrak{~~given~~}}}}$

in ∆ABC,AD is the median and E is the midpoint of AD that is ,is AE=DE and BE is produced to F that is BF intersect AC at F and BD=BC.

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$AF = \frac{1}{3} AC$

$\:\: \underline{\underline{\bf{\large\mathfrak{~~constrution~~}}}}$

• A line DG from the point D is drawn ||to BF.

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Now ..BF||DG

therefore...EF || DG. and BF||DG

now ....for the ∆ADG,,,

E is the midpoint of AD and EF is parallel to DG.

so from the converse mid point theorem we can say that it

F is the midpoint of AG.....

therefore...AF=FG

now again from the ∆ BFC,

D is the midpoint of BC and DG is parallel to BF.

so , again from the converse mid point theorem we can say that....

G is the midpoint of FC...

therefore...FG=GC

so.....AF=FG=GC

therefore...

=>AC=AF+FG+GC

=>AC=AF+AF+AF

=>AC=3AF

=>AF=(1/3)AF. (proved)

$\huge\mathcal\green{\underline{hope\:\: this\:\: helps\:\: you}}$

Answer 2

Answer:

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