PROOF OF PYTHAGORAS THEOREM
STATEMENT:- In a right angled triangle sum
of square of two sides of a triangle is equal to the square of the third
side.
Given:- In
triangle ABC, ∠B = 90o
To Prove :- AB2 + BC2 = AC2
Construction: Draw BD 丄 AC
Proof:- In△ADB and △ABC
∠1 = ∠3
∠A = ∠A
∴ By AA ～ rule
△ADB～△ABC
AB x AB =AC x AD
AB2 = AC x AD ................(1)
In△BCD and △ACB
∠2 = ∠3
∠C = ∠C
∴ By AA ～ rule
△BCD～△ACB
BC x BC =AC x CD
BC2 = AC x CD ..............(2)
Adding
equation (1) and (2)
AB2 + BC2 =
AC X AD + AC X CD
AB2 + BC2 = AC(AD+CD)
AB2 + BC2 = AC X
AC
AB2 + BC2 = AC2
Hence prove the required theorem
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PROOF OF CONVERSE OF PYTHAGORAS THEOREM
STATEMENT:-
If sum of squares of two sides of a
triangle is equal to the square of the third side then the angle opposite to
the larger side is right angle.
GIVEN :- In triangle ABC, AB2 +
BC2
= AC2
TO PROVE:- ∠B = 90o
CONSTRUCTION:-
Draw another △ DEF such that AB = DE, BC = EF, and ∠E = 90o
PROOF:-
In △DEF , ∠E
= 90o
Therefore by Pythagoras Theorem
DE2 + EF2 =
DF2
Putting AB = DE and BC = EF we get
AB2 + BC2 =
DF2 ..........................(1)
But AB2 + BC2 = AC2
..........................(2)Given
From (1) and (2) we get
AC2 = DF2
Or
AC = DF
Now In△ ABC and △DEF
AB = DE
..................(By construction)
BC = EF .....................(By
construction)
AC = DF
......................(Proved )
∴
By SSS ≌ rule
In △ABC ≌ △DEF
∠B = ∠E ...............(By CPCT)
But ∠E = 90o
∴ ∠B = 90o
Hence prove the required result
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