In quadrilateral ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that AP/PD = PC/BP

Question

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

Answer 1

proof: 

seg AD || seg BC and BD is their transversal. [Given] 

∴ ∠DBC ≅ ∠BDA [Alternate angles] 

∴ ∠PBC ≅ ∠PDA (i) [D – P – B] 

In ∆PBC and ∆PDA, 

∠PBC ≅ ∠PDA [From (i)] 

∠BPC ≅ ∠DPA [Vertically opposite angles] 

∴ ∆PBC ~ ∆PDA [AA test of similarity] 

∴ BP/PD = PC/AP  [Corresponding sides of similar triangles] 

∴ AP/PD = PC/BP  [By altemendo]