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 = 90^{o}
^{}
To Prove : AB^{2} + BC^{2} = AC^{2}
^{}
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
AB^{2} = AC x AD ................(1)
In△BCD and △ACB
∠2 = ∠3
∠C = ∠C
∴ By AA ～ rule
△BCD～△ACB
BC x BC =AC x CD
BC^{2} = AC x CD ..............(2)
Adding
equation (1) and (2)
AB^{2 }+ BC^{2} =
AC X AD + AC X CD
AB^{2 }+ BC^{2} = AC(AD+CD)
AB^{2 }+ BC^{2} = AC X
AC
AB^{2 }+ BC^{2} = AC^{2}
Hence prove the required theorem
*********************************************
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, AB^{2 }+
BC^{2}
= AC^{2}
^{}
TO PROVE: ∠B = 90^{o}
^{}
CONSTRUCTION:
Draw another △ DEF such that AB = DE, BC = EF, and ∠E = 90^{o}
^{}
PROOF:
In △DEF , ∠E
= 90^{o}
Therefore by Pythagoras Theorem
DE^{2 }+ EF^{2} =
DF^{2}
Putting AB = DE and BC = EF we get
AB^{2 }+ BC^{2} =
DF^{2 }..........................(1)
But AB^{2 }+ BC^{2} = AC^{2}
..........................(2)Given
From (1) and (2) we get
AC^{2} = DF^{2}
^{ }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 = 90^{o}
∴ ∠B = 90^{o}
^{}
Hence prove the required result
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