# <span lang="EN-US

To prove: AC bisects A and C, and AC is the perpendicular bisector of BD

Proof:

In ∆ABC and ∆ADC, we have

BC = DC … given

AC = AC … common side

Thus by SSS property of congruence,

Hence, we know that, corresponding parts of the congruent triangles are equal

BAC = DAC

BAO = DAO …(1)

It means that AC bisects BAD ie A

Also, BCA = DCA … cpct

It means that AC bisects BCD, ie C

BAO = DAO … from 1

AO = AO … common side

Thus by SAS property of congruence,

Hence, we know that, corresponding parts of the congruent triangles are equal

BOA = DAO

But BOA + DAO = 180°

2BOA = 180°

BOA = = 90°

So, BO = OD

Which means that AC = BD

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