27
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
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Sol.
PQ is diameter of the circle.
Therefore $\angle P'PQ = \angle Q'QP = 90^{\circ}$
$\angle PPQ + \angle Q'QP = 180^{\circ}$
$\Rightarrow$ PP' $||$ QQ'_a_1.png)
PQ is diameter of the circle.
Therefore $\angle P'PQ = \angle Q'QP = 90^{\circ}$
$\angle PPQ + \angle Q'QP = 180^{\circ}$
$\Rightarrow$ PP' $||$ QQ'