Question

hello everyone!!
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☺pls solve the above question no.7

thanks...

Answer 1

Consider ∆AOB & ∆AOD

<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

Answer 2

Here is your answer dear friend