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In ABD and BAC
AD=BC (given)
angleBAD=angleBAC(given)
AB=AB (common)
Triangle ABD CONGURENT TO triangleBAC
BY SAS CONGURENCY
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Given,
AD=BC
angle DAB = Angle CBA
i) in ∆ABD & ∆BAC
AD=BC
angle DAB = Angle CBA
AB=AB (COMMON)
∆ABD ~ ∆BAC (BY SAS)
ii) BD=AC (CPCT)
iii) ANGLE ABD = ANGLE BAC ( CPCT)
hope this may help
AD=BC
angle DAB = Angle CBA
i) in ∆ABD & ∆BAC
AD=BC
angle DAB = Angle CBA
AB=AB (COMMON)
∆ABD ~ ∆BAC (BY SAS)
ii) BD=AC (CPCT)
iii) ANGLE ABD = ANGLE BAC ( CPCT)
hope this may help
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