InterviewSolution
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InterviewSolution
| 1. |
In figure. 6.19 DE||ACand DF||AE.prove that BF/FE = BF/EC |
| Answer» In {tex}\\triangle ABE,{/tex} we have {tex}DE||AE,{/tex}then{tex}\\frac{{BD}}{{AD}} = \\frac{{BF}}{{FE}}\\,{/tex} [By\xa0BPT] ...... (i)In {tex}\\triangle ABC,{/tex}\xa0we have\xa0{tex}DE||AC,{/tex}\xa0then{tex}\\frac{{BD}}{{AD}} = \\frac{{BE}}{{EC}}\\,{/tex}\xa0[By\xa0BPT] ...... (ii)From (i) and (2), We get{tex}\\frac{{BF}}{{FE}} = \\frac{{BE}}{{EC}}{/tex} | |