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

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Answer 1

this is also known as basic proportionality theorem or Thales theorem
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Answer 2

Given that ;

∆PQR , in which XY || QR , and XY intersects PQ, and PR at X and Y respectively.

To prove :

$\frac{PX}{XQ}$ = $\frac{PY}{YR}$

Construction :

Join RX and QY and draw YN perpendicular to PX and QM perpendicular to PR.

[see the attachment]

Proof :

We know that -

Area of triangle = $\frac{1}{2}$ × Base × Height

Therefore,

Area of [∆PXY] = $\frac{1}{2}$ × PX × YN …(i)

Area of [∆PXY] = $\frac{1}{2}$ × PY × XM …(ii)

Area of [∆QXY] = $\frac{1}{2}$ × QX × YN …(iii)

Area of [∆RXY] = $\frac{1}{2}$ × YR × XM …(iv)

Now, dividing (i) by (iii), we get -

$\frac{ar(\triangle PXY)}{ar(\triangle QXY)}$

⇒ $\frac{\frac{1}{2}\times PX \times YN}{\frac{1}{2}\times QX \times YN}$

⇒$\frac{PX}{QX}$ … (v)

Again, dividing (ii) by (iv), we get -

$\frac{ar(\triangle PXY)}{ar(\triangle RXY)}$

⇒ $\frac{\frac{1}{2}\times PY \times XM}{\frac{1}{2}\times YR \times XM}$

⇒$\frac{PY}{YR}$ … (vi)

And, we know that ;

Area of triangles with same base and between the same parallels are equal. So,

Area of [∆QXY] = Area of [∆RXY] … (vii)

Therefore, from (v), (vi) and (vii), we get -

$\frac{PX}{XQ}$ = $\frac{PY}{YR}$

Hence, it is proved.