Question

Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct

points, the other two sides are divided in the same ratio.​

Answer 1

It is basic proportionality theorem

please mark it as a brainlist

Answer 2

BASIC PROPORTIONALITY THEOREM
OR,
THALES THEOREM


Given :- In triangle ABC, DE is parallel to BC

To prove :- AD/DB = AE/EC

Construction -

$join \: be \: and \: dc$

$dm \perp ae \\ \\ en \perp ad$

Proof :-

Area Triangle DBE and Area Triangle DEC are equal as they lie in same base and between same parallels.

......(i)

Area of Triangle AED is equal to :-

$= \frac{1}{2} \times dm \times ae \dots(ii) \\ \\ \\ = \frac{1}{2} \times en \times ad \dots(iii)$

Area of Triangle DEC is equal to :-

$= \frac{1}{2} \times dm \times ec$

Area of Triangle DEB is equal to :-

$= \frac{1}{2} \times en \times db$

Now,

From (i) we can say that :-

$\boxed{ \frac{ar(ade)}{ar(deb)} = \frac{ar(ade)}{ar(dec)} }$

Now,

Putting values :-

$\dfrac{ \frac{1}{2} \times en \times ad}{ \frac{1}{2} \times en \times db} = \dfrac{ \frac{1}{2} \times dm \times ae}{ \frac{1}{2} \times dm \times ec} \\ \\ \\ \\ \frac{ad}{db} = \frac{ae}{ec}$

Hence, proved!