Question

In the adjoining figure DE║AC. Prove that $\frac{BE}{EC} =\frac{BC}{CP}$

Answer 1

Given :- DE || AC & DC || AP

To Prove :- BE/EC = BC/CP

Solution :-

$In\:\triangleBCA,We\:have\:ED\:\parallelCA$

$\mathrm{By\:BPT\:Theorem,We\:get}$

$\implies\mathrm{\frac{BE}{EC}=\frac{BD}{DA}}-(1)$

$\mathrm{Similarly\:In\:\triangle\:ABP}$

$\implies\mathrm{\frac{BC}{CP}=\frac{BD}{DA}}-(2)$

$\textsf{From\:equation\:1\:and\:2\:We\:get}$

☆ BE /EC = BC/CP

Answer 2

Answer:

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