Question

In a figur ABP
In the figure DE || AC and DC || AP. Prove that BE/EC=BC/CP​

Answer 1

Use Basic Proportionality Theorem

Step-by-step explanation:

Given,

DE║AC

So, $\frac {BE}{EC} = \frac {BD}{DA}$ (Using Basic Proportionality Theorem)

• The Basic Proportionality Theorem states that if a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion.

Also, DC║AP

So, $\frac {BC}{CP} = \frac {BD}{DA}$ (Using Basic Proportionality Theorem)

Therefore,  

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

*Refer Figure Attached for Easier Understanding

Answer 2

Answer:

Step-by-step explanation:

In ∆BPA

DC || AP (given)

Therefore by basic proportionality theorem

BC/CP = BD/DA. -------(1)

In ∆BCA

DE||AC

Therefore by basic proportionality theorem

BE/EC = BD/DA ---------(2)

From (1)and (2)

BC/CP = BE/EC. Or. BE/EC = BC/CP

[Hence proved ]

Hope it will help you :-) :-)