Problem: In parallelogram ABCD of the accompanying diagram, line DP is drawn bisecting BC at N and meeting AB (extended) at P. From vertex C, line CQ is drawn bisecting side AD at M and meeting AB (extended) at Q. Lines DP and CQ meet at O. If the area of parallelogram ABCD is k, then the area of triangle QPO is equal to
Answer Choices:
A. k
B. 6k/5
C. 9k/8
D. 5k/4
E. 2k
Solution:
Since DM=AM,∠QMA=△DMC and ∠CDM=∠QAM we have △QAM≅△MCD.
Similarly △BPN≈△DNC. Now,
area △QPO= area ABCD+ area △DOC
and
area △DOC=41(21 area ABCD)=8k,
so that
area △QPO=k+8k=89k.