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ABCD is a quadrilateral such that AB = AD and CB = CD. Prove that AC is the perpendicular bisector of BD.

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Given: ABCD is quadrilateral

And AB = AD, CB = CD

To prove : AC is \perp^{r} bisector at BD

In \triangleABC and \triangleADC

AB = AD                     (Given)

BC = CD                     (Given)

AC = AC                     (Common)

\triangleABC \cong \triangleADC           (by SSS congruency)

Then by CPCT,

ÐBAC = ÐDAC          …(i)

Now in \triangleABO and \triangleADO

AB = AD                     (Given)

AO = AC                     (common)

\angleBAO = \angleDAO (from i)

\triangleABO \cong \triangleDAO          (by SAS congruency)

Then by CPCT,

\therefore \angleAOB = \angleDOA

\because\angleBOD = 180°

\angleAOB + \angleAOD = 180°                     (\because\angleBOD = \angleAOB + \angleAOD)

\angleAOB + \angleAOB = 180°                      (\because\triangleAOB \cong \triangleAOD)

\angleAOB = 90°                                       (\angleAOB = \angleAOD)

Hence, proved.

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