Consider two right triangles ABC and PQR in which\xa0{tex} \\angle B{/tex} \xa0and\xa0{tex}\\angle Q{/tex} are the right angles.We have,In\xa0{tex}\\triangle ABC{/tex}{tex}\\sin B=\\frac{AC}{AB}{/tex}\xa0and, In\xa0{tex}\\triangle PQR{/tex}\xa0{tex}\\sin Q=\\frac{PR}{PQ}{/tex}{tex} \\because \\quad \\sin B = \\sin Q{/tex}{tex} \\Rightarrow \\quad \\frac { A C } { A B } = \\frac { P R } { P Q }{/tex}{tex} \\Rightarrow \\quad \\frac { A C } { P R } = \\frac { A B } { P Q } = k{/tex}(say) …… (i){tex} \\Rightarrow {/tex} AC = kPR and AB = kPQ …..(ii)Using Pythagoras theorem in triangles ABC and PQR, we obtain\xa0AB2 = AC2 + BC2 and PQ2 = PR2 + QR2{tex} \\Rightarrow \\quad B C = \\sqrt { A B ^ { 2 } – A C ^ { 2 } } \\text { and } Q R = \\sqrt { P Q ^ { 2 } – P R ^ { 2 } }{/tex}{tex} \\Rightarrow \\quad \\frac { B C } { Q R } = \\frac { \\sqrt { A B ^ { 2 } – A C ^ { 2 } } } { \\sqrt { P Q ^ { 2 } – P R ^ { 2 } } } = \\frac { \\sqrt { k ^ { 2 } P Q ^ { 2 } – k ^ { 2 } P R ^ { 2 } } } { \\sqrt { P Q ^ { 2 } – P R ^ { 2 } } }{/tex}\xa0[ using (ii) ]{tex} \\Rightarrow \\quad \\frac { B C } { Q R } = \\frac { k \\sqrt { P Q ^ { 2 } – P R ^ { 2 } } } { \\sqrt { P Q ^ { 2 } – P R ^ { 2 } } } = k{/tex}…(iii)From (i) and (iii), we get{tex} \\frac { A C } { P R } = \\frac { A B } { P Q } = \\frac { B C } { Q R }{/tex}{tex} \\Rightarrow \\quad \\Delta A C B – \\Delta P R Q{/tex}\xa0[By S.A.S similarity]{tex} \\therefore \\quad \\angle B = \\angle Q{/tex}\xa0Hence proved.