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Question
In the following figure, the line ABCD is perpendicular to PQ; where P and Q are the centers of the circles.
Show that:
(i) AB = CD ;
(ii) AC = BD.
Solution
In the circle with center Q, QO ⊥ AD
∴ OA = OD ...(i) ...[perpendicular drawn the center of a circle to a chord bisects it]
In circle with center P, PO ⊥ BC
∴ OB = OC ....(ii) ....[perpendicular drawn the center of a circle to a chord bisects it]
(i) - (ii) gives,
AB = CD ...(iii)
(ii) Adding BC to both sides of equation (iii)
AB + BC + CD + BC
⇒ AC = BD
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