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consider triangle ACD and ABC
AB = AD (given)
‹DAC = ‹BAC (As AC bisects ‹A)
AC=AC (common)
TRIANGLE ACD = triangle ABC (BY SAS)
AB = AD (given)
‹DAC = ‹BAC (As AC bisects ‹A)
AC=AC (common)
TRIANGLE ACD = triangle ABC (BY SAS)
anusaji:
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