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Tangent Perpendicular to Radius at Point of Contact

In figure, O ...

Question

In figure, O is the centre of the circle and BCD is tangent to it at C. Prove that $∠ B A C + ∠ A C D = 90^{\circ}$

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Solution

OA = OC (radii of same circle)

OAC = OCA ...(1)

Since the tangents at any point of a circle is perpendicular to radius at the point of contact.

OCD = 90^o

ACD + OCA = 90^o

ACD + OAC = 90^o [From (1)]

ACD + BAC = 90^o

Hence proved.

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