Question

$\Large\bf{ Question :-}$

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

Solve it with full explanation..

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Answer 1

Step-by-step explanation:

Given:

• ABC is a triangle in which
• DE intersect AB at D and AC at E
• DE is also parallel to another line BC.

To Prove:

• AD/DB = AE/EC

Construction:

• Join BE & CD
• Draw two perpendiculars first EG on AD & second DF on AE.

Proof: As we know that

★ Area of ∆ = 1/2(Base)(Height) ★

In ∆ADE and ∆DEB

➙ ar(∆ADE) = 1/2 $\times$ AD $\times$ EG

➙ ar(∆DEB) = 1/2 $\times$ DB $\times$ EG

Therefore,

$\implies \: \frac{ar(ADE)}{ar(DEB) \: } \: = \frac{ \frac{1}{2} \times AD \times EG}{ \frac{1}{2} \times DB \times EG} \\ \\ \implies \: \frac{ar(ADE)}{ar(DEB)} = \frac{AD}{DB}......(1)$

Now, in ∆ADE & ∆CED

➙ ar(∆ADE) = 1/2 $\times$ AE $\times$ DF

➙ ar(∆CED) = 1/2 $\times$ EC $\times$ DF

Therefore,

$\implies \: \frac{ar(ADE)}{ar(CED)} = \frac{ \frac{1}{2} \times AE \times DF }{ \frac{1}{2} \times EC \times DF} \\ \\ \implies \: \frac{ar(ADE)}{ar(CED)} \: = \frac{AE}{EC} ......(2)$

Now, As we know that if two triangles are between same base and between same parallel then the area of triangles will be equal to each other.

Here, ∆DEC & ∆BDE are on the same base DE and between same parallels DE and BC therefore,

• ar(∆DEC) = ar(∆BDE)......(3)

Hence,

$= > \frac{ar(ADE)}{ar(BDE)} = \frac{ar(ADE)}{ar(DEC)} \\ \\ = > \frac{AD}{DB} = \frac{AE}{EC} \: \: \: from \: (1) \: and(2)$

Hence, Proved.

Answer 2

Simple question

DE is parallel to BC,

so,angle ABC=angle ADE and angle AED=angleACB

{Reason:-Alternate interior angle)

so,

∆ADE~∆ABC

Now using the property of similar triangle we can write

AB/AD=AC/AE (ratio of corresponding sides are eqaul)

subtract 1 from both side

AB/AD -1 =AC/AE -1

(AB-AD)/AD=(AC-AE)/AE

BD/AD=EC/AE

This shows that line DE which is parallel to third side divide the line in the fixed ratio