Proof of Pythagoras Theorem and its Converse
PYTHGORAS THEOREM AND ITS CONVERSE
CLASS 10 CHAPTER 6
Proof of Pythagoras theorem and its converse for class X, complete explanation of the Pythagoras theorem and its converse, Statement and proof of Pythagoras theorem class x, statement and proof of converse of Pythagoras theorem.
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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