Question

In fig. chord AB bisects chord CD at O, Prove that –
OA² = DO·OC

Answer 1

Given : chord AB bisects chord CD at O

CO = OD …..(i)

Join BC

∠CAB and ∠BDC, arc angles of same segment so will be equal

∠CAB = ∠BDC …..(ii)

In ∆COA and ∆DOB

∠CAO = ∠BDO (from equation (ii))

CO = OD (given eqn. (i)

∠COA = ∠DOB( vertically opposite angles)

By ASA congruence rule.

∆COA = ∆DOB

Thus, corresponding sides of congruent triangle are same

OA = OB

OB = OC [∵ OB and OC, arc radius of circle]

OA = OC

(OA)² = (OC)²

(OA)² = OC² × OC²

OA² =OC × OD [OC = OD]

or OA² = DO × OC