Question

in the given figure O is the centre of the circle and BCD is tangent to it at C proove that Angle BAC + Angle ACD = 90 degree.

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Answer 1

BCBC is tangent, we have

∠OCD=90

∠OCD=90

OA=OCOA=OC, so ∠OAC=∠OCA∠OAC=∠OCA. Hence

∠BAC+∠ACD=∠OCA+∠ACD=∠OCD=90

Answer 2

Given,

BCBC is tangent, we have

∠OCD=90°

∠OCD=90°

OA=OCOA=OC, so ∠OAC=∠OCA∠OAC=∠OCA. Hence

∠BAC+∠ACD=∠OCA+∠ACD=∠OCD=90°

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