Answer :

Given that,

PSDA is a parallelogram

Since,

AP ‖ BQ ‖ CR ‖ DS and AD ‖ PS

Therefore,

PQ = CD (i)

In

C is the mid-point of BD and CF ‖ BE

Therefore,

F is the mid-point of ED

EF = PE

Similarly,

PE = FD (ii)

In PQE and , we have

PE = FD

∠EPQ = ∠FDC (Alternate angle)

And,

PQ = CD

So, by SAS theorem, we have

Area (Δ PQE) = Area (Δ CFD)

Hence, proved

Rate this question :

How useful is this solution?

We strive to provide quality solutions. Please rate us to serve you better.

Related Videos

Division of Polynomials46 mins

Newton's Second Law46 mins

R D Sharma | Surface Area and Volumes39 mins

Surface Area and Volume of Right Circular Cone37 mins

Final Phase & Legacy of the Revolution43 mins

Newton's Third Law44 mins

Miscellaneous questions42 mins

Smart Revision | Surface Areas and Volume of Solids31 mins

Significance of Newton's Laws in daily life42 mins

Newton's First Law45 mins

Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Expertsview all courses

Dedicated counsellor for each student

24X7 Doubt Resolution

Daily Report Card

Detailed Performance Evaluation