Question

In given figure, 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  AOB = 90°.​

Answer 1

Answer:

ANSWER

Consider the problem

Let us join point O to C

In ΔOPAandΔOCA

OP=OC (Radii of the same circle)

AP=AC (Tangent from point A)

AO=AO (Common side)

ΔOPA≅ΔOCA (SSS congruence criterion)

Therefore, P↔C,A↔A,O↔O

∠POA=∠COA.........(1)

Similarly,  

∠QOB≅∠OCB

∠QOB=∠COB.........(2)

Since,POQ is a diameter of the circle, it is a straight line.

Therefore, ∠POA+∠COA+∠COB+∠QOB=180  

∘

 So, from equation (1) and equation (2)

2∠COA+2∠COB=180  

∘

 ∠COA+∠COB=90  

∘

∠AOB=90  

∘

 

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Step-by-step explanation: