In ▢ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that APPD=PCBP

In ▢ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect ea

1.	In ▢ABCD, seg AD \|\| seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that APPD=PCBP
Answer» In ▢ABCD, seg AD \|\| seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that $\frac{AP}{PD} = \frac{PC}{BP}$