Question

quadrilateral ABCP and parallelogram ABCD are on same base AB diogonal AC bisects the quadrilateral prove that PD parellel AC

Answer 1

Here quadrilateral ABCP and ABCD are on the same base
AC bisect ABCP (given)
ar. (∆ABC) = ar. (∆APC) .........(1)
AC bisect parallelogram ABCD (Diagonal of a parallelogram divides it into 2 triangles of equal areas)
=> ar. (∆ABC) = ar. (∆ACD)
So ar(ΔAPC) = ar(ΔACD) ........(2)
from (1) and (2), we get
And they are lying on the same base AC
And from the property if the area of two triangles lying on the same base have equal area then they are lying between two parallel lines