Question

State the basic proportionality theorem and also write its converse.

Please give proper answers.

Answer 1

Heya !

Basic Proportionality Theorem..

Statement..
If a line parallel to a side of a triangle and intercept the remaining sides in two distinct points , then the line divides the side in the same proportion.

Converse of basic Proportionality Theorem

If a line divides any two sides of a triangle in the same ratio , then the line is parallel to the third side.

Thanks!

#BeBrainly!!

Answer 2

$\huge \red{Hello \: Friend}$

$\green{Theorem :}$

If a line parallel to a side of triangle and intersects the remaining sides in two distinct points,then the line divides the sides in the same proportion.

$\orange{Lets \: prove \: it}$

$\green{Given : }$

In triangle ABC, seg DE ll seg BC

$\green{To \: prove : } \frac{AD}{DB} = \frac{AE}{EC}$

$\green{Construction : }$

Draw seg DC and seg BE and also seg DF and seg GE.

$\orange{Proof : }$

In triangle ADE and triangle DEB have equal heights.

$\frac{A(traingle \: ADE)}{A(triangle \: DEB)} = \frac{AD}{DB} .....1(areas \: proportionate \: to \: bases)$

Triangle ADE and triangle DEC have equal heights.

$\frac{A(triagle \: ADE)}{A(triangle \: DEC)} = \frac{AE}{EC} .....2(areas \: proportionate \: to \: bases)$

seg DE is common base of triangle DEB and triangle DEC.seg DE ll seg BC hence, triangle DEB and triangle DEC have equal heights.

$A(triangle \: DEB) = A(triangle \: DEC)...(3)$

$\frac{A(triangle \: ADE)}{A(triangle \: DEB)} = \frac{A(triangle \: ADE)}{A(trianle \: DEC)} ....(from \: (1).(2) \: and (3)$

$\frac{AD}{DB} = \frac{AE}{EC} ...(from \: (1) \: and \: (2)$

$\boxed{ \boxed{Hence \: proved}}$

$\blue {Converse \: of \: B.P.T}$

$\red{Theorem : }$

If a line divides any two sides of a triangle in the same ratio, then that line is parallel to the third sides.

$\orange{this \: theorem \: can \: \: be \:proved \: indirect \: methode }$

$\boxed{ \boxed {thanks}}$