Answer:
Given : Two 
ABC and
DBF such that 
To prove: 
Construction: Draw 
and 
Proof: 
?(1)
Again, in 
and 
, we
have
(Each = 90o)
(By A.A. rule)
?(2)
[
corresponding
sides of similar triangles are proportional]
Further, 
(Given)
?(3)
[ corresponding
sides of similar triangles are proportional]
From (2) and (3), 
Putting in (1), we get
of (2)]
Hence 
Q.E.D
OR
Given : A triangle ABC such that AC2 = AB2
+ BC2
To prove : 
Construction: Draw a 
such
that DE = AB, EF = BC and 
is
right angled at E
[By
construction]
AB2
+ BC2 = DF2 (By Pythagoras
Theorem)
But 
?(1)
[ DE =
AB, EF = BC]
(Given)
?(2)
Now, in 
and 
and
Hence 
i.e., Q.E.D
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