21u cant prove it because not always B =D=90° ( it must be given AD and BC are parallel)
draw figure in some other fashions u will get cases where A=C=90 B≠D≠90
In the above figure we see four line segments intersecting at
A,B,C and D respectively. LA = 900, LC = 900.
Surely, LB = LD = 900. So how can we prove it theoretically...!
Practically its true and obvious but can that be proved theoretically..?
(i'm confused)
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4 Answers
21
36@shubhodip -> Hi! dear.... there can be no such case that A=C=90 B≠D≠90......????
21how about that ?
actually quadrilaterals inscribed in a circle which has only one diagonal passing through the center of the circle are all of such form...
36gr8.... I was totally cofused and couldn't figure it out......!!!!!!!!!!!!!!.....
THANKS....