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Two tangents PA and PB are drawn to a circle with centre o from an external point p ..prove that

Answer» Suppose OP intersects AB at C.In triangles PAC and PBC, we havePA = PB\xa0[{tex} \\because{/tex}\xa0Tangents from an external point are equal]{tex}\\angle A P C = \\angle B P C{/tex}\xa0[{tex}\\because{/tex}\xa0PA and PB\xa0are equally inclined to\xa0O P]and, PC = PC [Common]So, by SAS-criterion of similarity, we obtain{tex}\\Delta P A C \\cong \\Delta P B C{/tex}{tex}\\Rightarrow{/tex}\xa0AC = BC and\xa0{tex}\\angle A C P = \\angle B C P{/tex}But, {tex}\\angle A C P + \\angle B C P{/tex}= 180°{tex}\\therefore \\quad \\angle A C P = \\angle B C P{/tex}= 90°Hence,\xa0{tex} O P \\perp A B{/tex}


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