Answer :

Let the external point be A.

AB and AC are the two tangents from A to the circle with radius O.



Given: AP and AQ are 2 tangents on the circle with radius O.


To Prove: AB = AC


Join, BO, AO and CO.


Proof:


Property: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.


By above property,


OBAB and OCAC


OBA = OCA = 90°


Now,


In ∆OAB and ∆OAC,


OB = OC [radius]


OBA = OCA = 90°


OA = OA [common]


By SAS, OBA OCA


AB = AC [By CPCTC]


Hence, Proved


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