in a parrallogram abcd , bc=ad and angle a is 90 than show it is a sequre.
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given:abcd is a parallelogram where bc=ad
angle a=90degree
rtp: abcd is a square
construction:db is joined
proof:in triangle adb &bcd
ad=bc
bd is the common arm
&abd=alternate angle bdc
so triangle abd congruent bcd(sas)
therefore,ab=dc(cpct)
so,in the parralelogram abcd the four sides are equal
therefore abcd is a square
THUS PROVED
angle a=90degree
rtp: abcd is a square
construction:db is joined
proof:in triangle adb &bcd
ad=bc
bd is the common arm
&abd=alternate angle bdc
so triangle abd congruent bcd(sas)
therefore,ab=dc(cpct)
so,in the parralelogram abcd the four sides are equal
therefore abcd is a square
THUS PROVED
sharanyamathi:
the question has to be quadrilateral
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Given:parallelogram abcd, a=90
to prove : abcd is a square
proof: in triangles abd and bcd,
ad=bc- s
cdb=dba-a(alternate angles)
ab=dc -s
by sas congruence rule triangle abd≈bcd.
since two pair of opposite sides are equal and one angle is 90, it is a square
to prove : abcd is a square
proof: in triangles abd and bcd,
ad=bc- s
cdb=dba-a(alternate angles)
ab=dc -s
by sas congruence rule triangle abd≈bcd.
since two pair of opposite sides are equal and one angle is 90, it is a square
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