Math Assignment Class VIII | Square & Square Root

Math Assignment  Class VIII | Square & Square Root Download or Print free  assignment with answer key  for   Class  8 Squares and  Square Roots.   Important and extra questions that cover all topics of square and square root and is useful and helpful for the students. Math Assignment  Class VIII | Square & Square Root LEVEL -1

Basic Proportionality Theorem (BPT), Thales Theorem

Basic Proportionality Theorem Class 10th
(OR)  B.P.T.   or Thales Theorem
Converse of basic proportionality theorem, thales theorem 10th standard, theorem 6.2 class 10

Statement:-
If a line is drawn parallel to one side of the triangle to intersect the other two sides in two distinct points, the other two sides are divided in the same ratio.
Given:-
A  Î” ABC in which line  BC, intersect side AB and AC at point D and E

To Prove :-  $\frac{AD}{DB}=\frac{AE}{EC}$
Construction :-
Draw  EM ä¸„ AB and DN ä¸„ AC. Also join BE and CD
Proof :- Area of triangle = $\frac{1}{2}$ ✕ Base ✕ Height
Area of △ ADE = $\frac{1}{2}$ ✕ AD ✕ EM ..............(1)

Area of △ BDE = $\frac{1}{2}$ ✕ BD ✕ EM ..............(2)

Divide equation (1) by equation (2) we get

$\frac{Ar(\Delta ADE)}{Ar(\Delta BDE)}=\frac{\frac{1}{2}\times AD\times EM}{\frac{1}{2}\times BD\times EM}$

$\frac{Ar(\Delta ADE)}{Ar(\Delta BDE)}=\frac{AD}{BD} \; \: \: .........\; (3)$
Similarly

$\frac{Ar(\Delta ADE)}{Ar(\Delta CDE)}=\frac{\frac{1}{2}\times AE\times DN}{\frac{1}{2}\times EC\times DN}$

$\frac{Ar(\Delta ADE)}{Ar(\Delta CDE)}=\frac{AE}{CE}\: \: \: \: ....................(4)$

Î”BED and Î”CDE are two triangles on the same base and lie between the same parallel DE and BC

Ar(Î”BED) = Ar(Î”CDE)   .................. (5)

From equation  (3), (4), (5) we get

$\frac{Ar(\Delta ADE)}{Ar(\Delta BED)}=\frac{Ar(\Delta ADE)}{Ar(\Delta CDE)}$

$\frac{AD}{DB}=\frac{AE}{EC}$

Hence prove the Basic Proportionality Theorem.

Note:- For the examination point of view students should study the basic proportionality theorem only its converse is only a motivational theorem.

Converse of Basic Proportionality Theorem

Statement:-

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

Given:-  In Triangle ABC,   $\frac{AD}{DB}=\frac{AE}{EC}$
To Prove :-  Line DE BC

Construction:-

If DE is not parallel to BC, then let us take another line DE' BC
Proof:-

In Î”ABC,  DE' BC Therefore by B.P.T

$\frac{AD}{DB}=\frac{AE'}{E'C}$

$But\: \frac{AD}{DB}=\frac{AE}{EC}\: \: ............Given$

Therefore  $\frac{AE}{EC}=\frac{AE'}{E'C}$

$1+\frac{AE}{EC}=1+\frac{AE'}{E'C}$

$\frac{EC+AE}{EC}=\frac{E'C+AE'}{E'C}$

$\frac{AC}{EC}=\frac{AC}{E'C}$

$EC =E'C$

This is possible only if E and E' coincide with each other

⇒ E and E' represent the same point on the side of the triangle.

Hence DE is parallel to the side BC

Converse of BPT is proved

Important result based on BPT

If a line intersects side AB and AC of a Î”ABC at D and E respectively and is parallel to BC, then prove that
$\frac{AD}{DB}=\frac{AE}{AC}$
Solution :-
It is given that DE॥BC, therefore by BPT

$\frac{AD}{DB}=\frac{AE}{EC}$

$\frac{DB}{AD}=\frac{EC}{AE}$   .........  (∵  By Invertendo

Adding 1 on both side we get

$\frac{DB}{AD}+1=\frac{EC}{AE}+1$

$\frac{DB+AD}{AD}=\frac{EC+AE}{AE}$

$\frac{AB}{AD}=\frac{AC}{AE}$

$\frac{AD}{AB}=\frac{AE}{AC}$     ..........  ( ∵  By Invertendo