Question

in the adjoining DE||AC and BE/EC=BC/CP. Prove that DC||AP.

Answer 1

I hope you understood My handwriting

Answer 2

Step-by-step explanation:

$In\:\triangle BPA, \\ we \:have\: DC \parallel AP$

/* By Basic proportionality theorem:

$\frac{BC}{CP}=\frac{BD}{DA}\: ----(1)$

In ∆BCA, we have

DE\\AC ( given )

$\frac{BE}{EC}=\frac{BD}{DA}\:---(2)$

/* By Basic proportionality theorem*/

From (1) and (2), we get

$\frac{BC}{CP}=\frac{BE}{EC}\\Or \\\frac{BE}{EC}=\frac{BC}{CP}$

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