Answer:
Ist Part
Similarly,
But
(Triangles on same base DE and between same parallells DE,
BC)
IInd Part
Draw AC, O is intersection of AC and EF. In and , EO ||
DC
?(i)
In and
By (i) and (ii),
Hence Proved.
OR
First Part Book Work
Second Part
We have,
AC = BC and AB2 = 2AC2
Now, AB2 = 2AC2 AB2
= AC2 + AC2
AB2 = AC2 + BC2
is a
right triangle right-angled at C
[(Given)]
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