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°.

Answer 1

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.