In `square ABCD`, seg `AD||` seg BC. Diagonal AC and digonal BD intersect each other in point P. Then show that `(AP)/(PD)=(PC)/(BP)`.

Explanation

seg AD`||` seg BC and line DB is transversal,
`/_ADP=/_CBP` …….(Alternalte angles )…………1
In `DeltaADP` and `DeltaCBP`,
`/_ADP~=/_CBP`….[From 1]
`/_APD~=/_CPB`………..(Vertically opposite angles)
`:.DeltaADP~DeltaCBP`………..(AA test of similarity)
`:.(AP)/(CP)=(PD)/(BP)`......(Corresponding sides of similar triangles are in proportion)
`:.(AP)/(PD)=(CP)/(BP)` ...........(By alternendo)
i.e. `(AP)/(PD)=(PC)/(BP)`