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In figure. 6.19 DE||ACand DF||AE.prove that …

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In figure. 6.19 DE||ACand DF||AE.prove that BF/FE = BF/EC

Posted by Nirankar Sharma 6 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 3 months ago

In {tex}\triangle ABE,{/tex} we have {tex}DE||AE,{/tex}then
{tex}\frac{{BD}}{{AD}} = \frac{{BF}}{{FE}}\,{/tex} [By BPT] ...... (i)
In {tex}\triangle ABC,{/tex} we have {tex}DE||AC,{/tex} then
{tex}\frac{{BD}}{{AD}} = \frac{{BE}}{{EC}}\,{/tex} [By BPT] ...... (ii)
From (i) and (2), We get
{tex}\frac{{BF}}{{FE}} = \frac{{BE}}{{EC}}{/tex}

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CBSE > Class 10 > Mathematics