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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