solve no 3. . . . . . . . . . . .
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In the parallelogram ABCD
join AC
Now in ∆s ADC and ABC
AB=DC [ oppo. sides of a ||gm are equal ]
DA = BC
AC = AC [ common ]
so, ∆ABC ≈ ∆ADC
so, angle ADC = angle ABC
Hence proved.
join AC
Now in ∆s ADC and ABC
AB=DC [ oppo. sides of a ||gm are equal ]
DA = BC
AC = AC [ common ]
so, ∆ABC ≈ ∆ADC
so, angle ADC = angle ABC
Hence proved.
Answered by
1
hope it will help you
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