Given In quadrilateral ABCD ,AB=AD and CB =CD
Construction join AC and BD
To prove AC in the perpendicular bisector of BD
Proof In `Delta ABC and Delta ADC `
AB=AD [given ]
BC =CD [given]
and AC =AC [common side ]
`therefore Delta ABC cong Delta ADC ` [by SSS congruence rule ]
` rArr angle 1 = angle 2` [by CPCT ]
Now in `Delta AOB and Delta AOD ` AB= AD [given ]
`rArr angle 1 = angle 2` [ proved above ]
and AO=AO [common side]
`therefore DeltaAOB cong Delta AOD ` [by SAS congruence rule ]
`rarr BO=DO ` [by CPCT]
and `angle 3 = angle 4` [by CPCT ...(i)]
But `angle 3+angle 4 =180^(@)` [linear pair axiom ]
`rArr angle 3 +angle 3 =180^(@)` [from Eq. ...(i)]
`rArr 2angle 3 =180^(@)`
`rArr angle 3 =(180^(@))/(2) `
`therefore angle 3 = 90^(@)`
Ac is perpendicular bisector of BD .
