OR
State and prove converse of Pythagoras theorem and hence
show that in an isosceles triangle ABC with AC = BC and AB2 = 2AC2,
prove thatAnswer:
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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