plz..help me friends
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Hey there!!
In fig ,
in ∆ABD and ∆BCD
AB = CD (Given)
Ang ABD = AngCDB ( each of 90°)
BD = BD ( common)
Hence
∆ ABD. ~= ∆BCD ( SAS)
nd hence
AD = BC ( CPCT )
Hope this helps u
^_^
In fig ,
in ∆ABD and ∆BCD
AB = CD (Given)
Ang ABD = AngCDB ( each of 90°)
BD = BD ( common)
Hence
∆ ABD. ~= ∆BCD ( SAS)
nd hence
AD = BC ( CPCT )
Hope this helps u
^_^
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