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| 1. |
In the given figure PA,QB and RC aer perpendicular to AC. Prove that 1/x+/1z=1/y. |
| Answer» In the given figure we have\xa0{tex}P A \\perp A C{/tex}\xa0and\xa0{tex}Q B \\perp A C{/tex}.{tex}\\Rightarrow Q B \\| P A{/tex}In {tex}\\triangle PAC{/tex}\xa0and\xa0{tex}\\triangle Q B C{/tex}, we have\xa0{tex}\\angle QCB= \\angle PCA{/tex} ( Common )\xa0\xa0{tex}\\angle QBC= \\angle PAC{/tex} ( both are 90o\xa0).So by AA similarity rule ,\xa0{tex}\\triangle Q B C \\sim \\triangle P A C{/tex}.{tex}\\therefore \\frac { Q B } { P A } = \\frac { B C } { A C }{/tex}{tex}\\Rightarrow \\frac { z } { x } = \\frac { b } { a + b }{/tex}. .....................................(i) [by the property of similar triangles]In\xa0{tex}\\triangle RAC{/tex} ,\xa0{tex}Q B \\| R C{/tex}.So,\xa0{tex}\\triangle Q B A \\sim \\triangle R C A{/tex}.{tex}\\therefore \\frac { Q B } { R C } = \\frac { A B } { A C }{/tex}{tex}\\Rightarrow \\frac { z } { y } = \\frac { a } { a + b }{/tex}. .....................................(ii) [by the property of similar triangles]Form (i) and (ii), we obtain{tex}\\frac { z } { x } + \\frac { z } { y }{/tex}{tex}= \\left( \\frac { b } { a + b } + \\frac { a } { a + b } \\right) = 1{/tex}{tex}\\Rightarrow \\quad \\frac { z } { x } + \\frac { z } { y } = 1{/tex}{tex}\\Rightarrow \\frac { 1 } { x } + \\frac { 1 } { y } = \\frac { 1 } { z }{/tex}or {tex}\\frac { 1 } { x } + \\frac { 1 } { y } = \\frac { 1 } { z }{/tex}.Hence proved. | |