Question

STATE AND PROVE BPT AND PYTHAGORAS THEOREM

Answer 1

Both Pythagoras and Basic Proportionality theorem are here

Answer 2

BASIC PROPORTIONALITY THEOREM

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

Given :- In ∆ABC in which a line parallel to Side BC intersect other two sides AB & AC at points D& E respectively.

To Prove:- AD/DB = AE/EC

Construction:- Join BE & CD and draw DM Perpendicular to AC & EN perpendicular to AB.

To Proof :- ar(∆ADE) = 1/2 × AD× EN ______eq(1)

at(∆BDE) = 1/2 ×DB× EN ________eq(2)

Dividing the equations (1) & (2)

ar(∆ADE) = 1/2 × AD × EN / ar(∆BDE) = 1/2 ×DB×EN

=> ar(∆ADE) = AD / ar(∆BDE)=DB _______eq(5)

Similarly, ar(∆ADE)=1/2×AE×DM _________eq(3)

ar(∆DEC) = 1/2 ×EC × DM ________eq ( 4)

Dividing the equations (3) & (4)

ar(∆ADE)=1/2 × AE × DM / ar(∆DEC) = 1/2 ×EC× DM

=> ar(∆ADE) = AE/ar(∆DEC) =BC ________eq(6)

∆BDE & ∆DEC are on the same base & lie between the same parallels BC & DE.

so , ar (∆BDE) = ar(∆DEC)

From eqn (5) & (6)

AD/DB = AE/EC

Now Pythagoras theorem

In a Right Triangle the square of the hypotenuse is equal to the sum of the Squares of other two sides.

Given :-

• A right triangle ABC
• It is a right angle at B

To Prove:-

AC² = AB² + BC²

Construction:-

Draw BC perpendicular to AD

To Proof:-

In ∆ADB & ∆ABC

• Angle A = Angle A --( Common Angles)
• ADB = ABC --(Each 90°)
• ∆ADB = ∆ABC --(AA similarity)

=> AD/AB = AB/BC -- (By Corresponding sides of similar ∆'s)

=> AB² = AD × AC ______(1)

Now In ∆ CDB & CBA

• Angle C = Angle C ( Common Angles)
• CDB = CBA ( each 90°

=> By AA Similarity∆ CDB = CBA

CD / CB = CB/CA (By Corresponding sides of similar Triangles)

=> BC = CD × CA______(2)

Adding eqn (1) &(2)

AB + BC = AD × AC + CD × AC

AC(AC × AD)+AD

AC²

=> AB² + BC² = AC²