In Figure, `O` is the centre of the circle and `B C D` is tangent to it at `Cdot` Prove that `/_B A C+/_A C D=90^0` .

Explanation

To prove that ∠BAC + ∠ACD = 90°, we will follow these steps:

### Step 1: Identify the elements of the circle
Let O be the center of the circle, and let BCD be the tangent to the circle at point C.

### Step 2: Establish that OC is perpendicular to BCD
Since BCD is a tangent to the circle at point C, we know that the radius OC is perpendicular to the tangent line BCD at point C. Therefore, we can write:
\[ \angle OCD = 90^\circ \]
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