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| 1. |
pin triangle abc de paralel to bc find value of x |
| Answer» Given in triangle, {tex}{/tex}ABC DE\xa0{tex}\\|{/tex}\xa0BC\xa0{tex}\\frac { \\mathrm { AD } } { \\mathrm { DB } } = \\frac { \\mathrm { AE } } { \\mathrm { EC } }{/tex}\xa0(by BPT){tex}\\Rightarrow \\quad \\frac { x } { x - 2 } = \\frac { x + 2 } { x - 1 }{/tex}{tex}\\Rightarrow{/tex}\xa0x(x - 1) = (x + 2)(x - 2){tex}\\Rightarrow{/tex}\xa0x2\xa0- x = x2\xa0- 22\xa0{tex}\\Rightarrow{/tex}\xa0x2\xa0- x = x2\xa0- 4{tex}\\Rightarrow{/tex} x = 4 | |