XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. prove that
rudeawakening:
<AOB = 90 deegrees
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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
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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