Question

Xy and pq are two parallel tangents to a circle with centre o and another tension a b with point of contact c intersecting xy at a and p q h b prove that angle a or b is equals to 90 degree

Answer 1

GIVEN- a circle with centre o to which XY and XY' are tangents

TP- ∠AOB=90°

CONST.- join OY, OY' and OC

PROOF- In ΔOYA and ΔOCA

             OY = OC [radii]

             OA=OA [common]

            AY = AC [tangents]

⇒By SSS

    ΔOYA≈ΔOCA

⇒∠OY'A=∠OCA   [CPCT]---------------(1)

║ly ΔOY'B≈ΔOCB

⇒∠Y'BO=∠CBO   [CPCT]--------------------(2)

∠YAB+∠Y'BA=180°[co-interior angles]

2∠OAB+2∠OBA=180°      -------[from (1) and (2)]

2(∠OAB+∠OBA)=180°

∠OAB+∠OBA=90°---------------------(3)

In ΔAOB

∠OAB+∠OBA+∠AOB=180°   [Angle sum property of a triangle]

90°+∠AOB=180°

∠AOB=180°- 90°

∠AOB=90°

∴ Hence Proved

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