Question

ABCD is a ||gm. Any line through A cuts segment DC at P and segment BC produced at Q.
Prove that ar(∆BCP) = ar (∆DPQ)​

Answer 1

Step-by-step explanation:

Construction :join A and C

proof: BC extends upto point Q

therfore CQllAD

∆ACQ and ∆DQC are on the same base CQ and b/w the same parallels CQllAD ,so they are equal in ar(∆ACQ)-ar(∆CPQ)=ar(∆CPQ)

=> ar(∆APC)=ar(∆DPQ)-(i)

since,ABllDC (:: opposite sides Of a llgm are equal)

::∆APC and ∆BCP are on the same base pc and b/w the same ll , pcllab, so they are equal in area.

ar (∆BCP)=ar (∆BCP)- (ii)

from (i) and (ii)

ar (∆BCP)= ar(∆DCQ)