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In the given figure, AB is a chord of circle with centre ‘O’. CD is the diameter perpendicular to AB. Show that AD = BD. |
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Answer» CD is diameter, O is the centre. CD ⊥ AB; Let M be the point of inter-section. Now in ΔAMD and ΔBMD AM = BM [ ∵ radius perpendicular to a chord bisects it] ∠AMD =∠BMD [given] DM = DM (common) ∴ ΔAMD ≅ ΔBMD ⇒ AD = BD [C.P.C.T] |
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