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| 1. |
In the given figure D is a point on AC such that AD=2CD rFkk DE || AB? |
| Answer» Given AD = 2CDSo AC = AD + DC = 2CD + CD = 3CDIn {tex}\\triangle{/tex}CDE and {tex}\\triangle{/tex}CAB,{tex}\\angle{/tex}C = {tex}\\angle{/tex}C (Common)and {tex}DE\\|AB{/tex}So {tex}\\angle{/tex}CDE = {tex}\\angle{/tex}CAB(Corresponding angles){tex}\\therefore{/tex}\xa0{tex}\\triangle{/tex}CDE\xa0{tex}\\sim{/tex}\xa0{tex}\\triangle{/tex}CAB (By AA similarity rule)Now,\xa0{tex}\\frac { \\operatorname { ar } ( \\triangle D C E ) } { ar( \\triangle A C B ) }{/tex},\xa0{tex}{/tex}={tex}\\frac{CD^2}{AC^2}=\\frac{CD^2}{(3CD)^2}=\\frac{1}{9}{/tex} | |