Question

In the given figure, AB is a chord of circle with centre ‘O’. CD is the diameter perpendicular to AB. Show that AD = BD.

Answer 1

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]