Question

In the figure segment PQ is the diameter of

the circle with center O. The tangent to the

tangent circle drawn from point C on it ,

intersects the tangents drawn from points P

and Q at points A and B respectively ,

prove that ∠AOC = 90°​

Answer 1

Answer:

Refer attached image for your answer.

Answer 2

Solution :-

Construction :- Join OC .

In ∆POA and ∆COA , we have,

→ PO = CO { Radius.}

→ OA = OA { Common.}

→ AP = AC { Tangents from same external point to the circle are equal .}

so,

→ ∆POA ≅ ∆COA .{ By SSS congruence .}

then,

→ ∠POA = ∠COA = Let x . { By CPCT.}

Similarly,

In ∆QOB and ∆COB , we have,

→ QO = CO { Radius.}

→ OB = OB { Common.}

→ BQ = BC { Tangents from same external point to the circle are equal .}

so,

→ ∆QOB ≅ ∆COB .{ By SSS congruence .}

then,

→ ∠QOB = ∠COB = Let y . { By CPCT.}

since , PQ is the diameter of circle. PQ is a straight line.

therefore,

→ ∠POA + ∠COA + ∠QOB + ∠COB = 180°

→ x + x + y + y = 180°

→ 2(x + y) = 180°

→ x + y = 90° .

hence,

→ ∠BOC = (x + y) = 90° (Proved).

[ Note :- In question we have to prove ∠BOC = 90° , not ∠AOC = 90° .]

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