Question

plz plz plz explain this theorem​

Answer 1

Step-by-step explanation:

Basic Proportionality Theorem:-

Statement:-

In a triangle, if line is drawn parallel to one side of the triangle, then it divides other two sides in the same ratio.

$\rm Given:- \: \: In \: \triangle ABC \: \: PQ \parallel AC$

$\rm To\: Prove:- \: \: \dfrac{AP}{BP} = \dfrac{QC}{BQ}$

$\rm Construction:- \: \: QN \perp AB \: and \: PM \perp BC \: \\ (See \: 2nd \: attachment)$

Proof:-

As area of triangle = ½ × Base × Height

Consider ∆BPQ and ∆APQ

$\rm \dfrac{Area \: of \: \triangle BPQ}{Area \: of \: \triangle APQ} = \dfrac{ \dfrac{1}{2} \times BP \times NQ}{ \dfrac{1}{2} \times AP \times NQ} = \dfrac{AP}{BP}......(i)$

Consider ∆BPQ and ∆CPQ

$\rm \dfrac{Area \: of \: \triangle BPQ}{Area \: of \: \triangle CPQ} = \dfrac{ \dfrac{1}{2} \times BQ \times PM}{ \dfrac{1}{2} \times CQ \times PM} = \dfrac{BQ}{CQ}......(ii)$

Since, ∆APQ and ∆CQP are having same base and lies between parallel lines DE and BC.

$\rm Area \: of \: \triangle APQ = Area \: of \: \triangle CQP$

Taking reciprocals

$\rm \dfrac{1}{Area \: of \: \triangle APQ} = \dfrac{1}{Area \: of \: \triangle CQP}$

Multiplication the equation with Area of ∆BPQ

$\rm \dfrac{Area\: of \: \triangle BPQ}{Area \: of \: \triangle APQ} = \dfrac{Area\: of \: \triangle BPQ}{Area \: of \: \triangle CQP}$

$\rm \dfrac{AP}{BP} = \dfrac{QC}{BQ} \: \: \: \: \: (From \: (i) \: and (ii))$

$\boxed{ \rm \therefore \frac{AP}{BP} = \frac{QC}{BQ} }$

Hence Proved