\xa0Let AB be a chord of a circle with centre O, and let AP and BP be the tangents at A and B respectively. Let the two tangents AP and BP meets at P.\xa0Now Join OP. Suppose OP meets AB at C and the chord AB=AC+BC.We have to prove that\xa0{tex}\\angle P A C = \\angle P B C{/tex}In two triangles PCA and PCB, we havePA = PB\xa0[{tex} \\because{/tex}Tangents from an external point are equal]{tex}\\angle A P C = \\angle B P C{/tex}\xa0[{tex} \\because {/tex}\xa0PA and\xa0PB are equally inclined to OP]and, PC = PC [Common]So, by SAS-criterion of congruence, we obtain{tex}\\Delta P A C \\cong \\Delta P B C{/tex}{tex}\\Rightarrow \\quad \\angle P A C = \\angle P B C{/tex}