3 Tutor System
Starting just at 265/hour

# In figure, XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that $$∠ \ AOB \ = \ 90°$$.

Since tangents drawn from an external point to a circle are equal,

$$\because$$ $$AP \ = \ AC$$.
Thus in $$∆ \ PAO$$ and $$∠ \ AOC$$, we have

$$AP \ = \ AC$$
$$AO \ = \ AO$$ [Common]
$$PO \ = \ OC$$ [Radii of the same circle]
$$∆ \ PAO \ \cong \ AOC$$ [ By SSS-criterion of congruence]

$$∠ \ PAO \ = \ ∠ \ CAO \$$
$$\Rightarrow \ ∠ \ PAC \ = \ 2 ∠ CAO$$

Similarly, we can prove that

$$∠ \ QBO \ = \ ∠ \ CBO \$$
$$\Rightarrow \ ∠ \ CBQ \ = \ 2 ∠ \ CBO$$

Now, $$∠ \ PAC \ + \ ∠ \ CBQ \ = \ 180°$$
[Sum of the interior angles on the same side of transversal is 180°]

$$\Rightarrow 2 ∠ \ CAO \ + \ 2 ∠ \ CBO \ = \ 180°$$

$$\Rightarrow ∠ \ CAO \ + \ ∠ \ CBO \ = \ 90°$$
$$\Rightarrow 180° \ - \ ∠ \ AOB \ = \ 90°$$
[$$\because$$ $$∠ \ CAO \ + \ ∠ \ CBO \ + \ ∠ \ AOB \ = \ 180°$$]

$$\Rightarrow \ ∠ \ AOB \ = \ 90°$$