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