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☺pls solve the above question no.7
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<AOB = <AOD = 90°
AB = AD [ All Sides of a rhombus are equal ]
OB = OD [ Diagonal Bisects each other ]
By RHS congruency
∆AOB ≈ ∆AOD
Therefore < BAO = < DAO
i.e. AC Bisects < A
Take ∆ COD & ∆ COB
CD = BC
< COD = < COB
OD = BO
By RHS congruency
∆COD ≈ ∆ COB
< DCO = < BCO
i.e AC Bisects < C
Similarly ,
Take ∆ BOA & BOC
You will get < ABO = < CBO
and then take ∆ AOD & COD
You will get < ADO = < CDO
Therefore BD Bisects both < B & < D
Hence proved