Question

Prove that if two tangents are drawn to a circle from a point outside it, then the line
segments joining the point of contact and the exterior point are equal and they
subtend equal angles at the centre. ​

Answer 1

Answer:

In ΔAPO and ΔBPO

In ΔAPO and ΔBPOOA=OB (radii)

In ΔAPO and ΔBPOOA=OB (radii) AP=BP (theorem)

In ΔAPO and ΔBPOOA=OB (radii) AP=BP (theorem) OP=OP (common)

In ΔAPO and ΔBPOOA=OB (radii) AP=BP (theorem) OP=OP (common) ΔAPO≅ΔBPO (by SSS congruency)

In ΔAPO and ΔBPOOA=OB (radii) AP=BP (theorem) OP=OP (common) ΔAPO≅ΔBPO (by SSS congruency)by cpct,

In ΔAPO and ΔBPOOA=OB (radii) AP=BP (theorem) OP=OP (common) ΔAPO≅ΔBPO (by SSS congruency)by cpct,∠AOP=∠BOP (hence, they suntend equal angle at centre P)

In ΔAPO and ΔBPOOA=OB (radii) AP=BP (theorem) OP=OP (common) ΔAPO≅ΔBPO (by SSS congruency)by cpct,∠AOP=∠BOP (hence, they suntend equal angle at centre P) ∠APO=∠BPO (hence, they are equally inclined)

Step-by-step explanation:

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Answer 2

Answer:

∠APO=∠BPO (hence, they are equally inclined)

Step-by-step explanation:

In ΔAPO and ΔBPO

OA=OB (radii)

AP=BP (theorem)

OP=OP (common)

ΔAPO≅ΔBPO (by SSS congruency)

by cpct,

∠AOP=∠BOP (hence, they suntend equal angle at centre P)

∠APO=∠BPO (hence, they are equally inclined)