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Question
In the given figure, O is the centre of the circle and BCD is tangent to it at C. Prove that ∠BAC + ∠ACD = 90°.
Solution
In the given figure, let us join D an A.
Consider `ΔOCA`. We have,
OC = OA (Radii of the same circle)
We know that angles opposite to equal sides of a triangle will be equal. Therefore,
`∠OCA =∠OAC ` … (1)
It is clear from the figure that
`∠DCA +∠OCA=∠OCD`
Now from (1)
`∠DCA +∠OAC =∠OCD`
Now as BD is tangent therefore, `∠OCD=90^circ.`
Therefore `∠DCA +∠OAC = 90^circ`
From the figure we can see that `∠OAC =∠BAC`
`∠DAC + ∠BAC = 90^o`
Thus, we have proved.
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