In a quadrilateral ABCD, AB=AD and CB=CD . Prove that : AC is perpendicular bisector of BD.
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Answered by
7
as all the sides are equal and parallel so it is a square
in abo and cbo
ab=bc
ob (common)
oc = oa ( diagonals bisect each other)
by sss
abo cong. to cbo
/_ aob = /_cob (cpct)
/_ aob + /_cob =180° (L.P.A.)
/_aob +/_ aob =180°
2 /_ aob =180°
/_aob=90°
•°• Ac_|_ bd
in abo and cbo
ab=bc
ob (common)
oc = oa ( diagonals bisect each other)
by sss
abo cong. to cbo
/_ aob = /_cob (cpct)
/_ aob + /_cob =180° (L.P.A.)
/_aob +/_ aob =180°
2 /_ aob =180°
/_aob=90°
•°• Ac_|_ bd
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Answered by
30
<AOB +<AOB = 180
2<AOB =180
<AOB=<AOD=90 ------------------(3)
so from equation 1 and 3 we get
AC is the perpendicular bisector of BD
(hence proved
2<AOB =180
<AOB=<AOD=90 ------------------(3)
so from equation 1 and 3 we get
AC is the perpendicular bisector of BD
(hence proved
Attachments:
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