Given a ∆ABD in which AB=AD and AC bisects BD prove that :. ∆ABC=~∆ADC
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in ∆ abc and ∆ adc
ab = ad (given)
ac = dc (given)
ac = ac (common)
so, ∆abc =~ ∆adc (by sss)
ab = ad (given)
ac = dc (given)
ac = ac (common)
so, ∆abc =~ ∆adc (by sss)
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oyo:
r u sure that answer is right
thats not the answer of ur question
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