| In the given figure of \[\Delta ABC\],\[DE||AC\]. If\[DC||AP\], where point P lies on BC produced, then \[\frac{BE}{EC}\]= |
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A) \[\frac{BD}{CP}\]
B) \[\frac{BC}{CP}\]
C) \[\frac{BC}{DA}\]
D) None of these
Correct Answer: B
Solution :
Given, in \[\Delta ABC,\,\,DE\,|\,\,|\,\,AC\] So, \[\frac{BE}{EC}=\frac{BD}{DA}\] (i) [by basic proportionality theorem] Also, in \[\Delta ABP\], \[DC\,|\,|\,AP\] [given] So, \[\frac{BC}{EC}=\frac{BC}{CP}\] (ii) [by basic proportionality theorem] From Eqs. (i) and (ii), we get \[\frac{BE}{EC}=\frac{BC}{CP}\]
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