Question

If a line draw parallel to one side of the triangle to intersect other two sides in distinct points, the other two sides are divided in the same ratio.​

Answer 1

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Step-by-step explanation:

➣Given: ∆ABC is which a line Parrellel two side BC intersect other two side at AB and AC at D and E.

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$\Large\bold{Prove} : \: \large \bold{ \frac{AD}{DB} = \frac{AE}{EC} }$

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$\: \bold{Proof: \: Let\: us \:join\: BE\: and\: CD \:and\: then\: Draw \:DM\:\perp\:AC \: and \: EN\:\perp\:AB.}\:$

$\rightarrow\: \mathtt{Area \: of \: \triangle \: ADE} = \mathtt{ \frac{1}{2} \times b \times h} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \mathtt{ \frac{1}{2} \times AD \times EN}$

$\rightarrow \mathtt{Area \: of \: \triangle \: BDE}\: = \mathtt{\frac{1}{2} \times b \times h} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \mathtt{\frac{1}{2} \times BD×EN}$

$\rightarrow \large \mathtt{ \frac{Area \: of \: \triangle{ADE} }{ Area \: of \: \triangle BDE }} = \mathtt{ \frac{ \cancel{\frac{1}{2}} \times AD \times \cancel{EN} }{ \cancel{\frac{1}{2}} \times BD \times \cancel{EN} }} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \Large\fbox\mathtt{ \frac{ AD }{ EN } }$

$\rightarrow \large\mathtt{Area of \triangle{ADE }}= \large\mathtt{\frac{1}{2} \times b \times h} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \large\mathtt{ \frac{1}{2} \times AE \times DM}$

$\rightarrow \: \large\mathtt{Area \: of \: \triangle{DEC}} = \mathtt{ \frac{1}{2} \times b \times h} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \large \mathtt{\frac{1}{2} \times EC \times DM}$

$\rightarrow \: \large\mathtt{ \frac{Area \: of \: \triangle ADE}{Area \: of \: \triangle \: DEC} } = \large \mathtt{ \frac{ \cancel{\frac{1}{2}}\times AE \: \times \cancel{DM} }{ \cancel{\frac{1}{2}} \times EC \times \cancel{DM} } } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \fbox\mathtt{ \frac{AE}{EC}}$

$\large \fbox \bold{Hence \: proved = \frac{ AD }{BD}= \frac{ AE}{ EC}}$

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