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In the given figure, O is the centre of the circle and *BCD *is tangent to it at *C*. Prove that ∠*BAC* + ∠*ACD* = 90°.

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#### Solution

In the given figure, let us join *D* an *A*.

Consider `Δ OCA`. 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^o`

Therefore `∠ DCA +∠ OAC = 90^o`

From the figure we can see that `∠ OAC =∠ BAC`

`∠ DAC + ∠ BAC = 90^o`

Thus we have proved.

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