Quadrilateral ABCD is a rhombus and AB is extended to P such that BP=AB . Prove that triangles DAB and CBP are congruent and diagonal BD is parallel to PC .

Proof : AB=AD=BC [Quadrilateral ABCD is a rhombus] angle B=angle D Which angle B? Which angle D? angle p=angle C [Both triangles ABD and PBC are isosceles triangles] Which angle C? angle B+angle D = angle P + angle c This is very confused! angle A=180-angle B-angle D angle B=180-angle P-angle C angle A=angle B No. Hence , triangle DAB is congruent to triangle ABD This is the same triangle! Since angle P = angle ABD [corresponding angle] , BD // PC The straight line that passes through D , parallel to AC meets PC prodiced at R , Prove that . I am not really sure bout this part. I cant see that which triangle is congruent to which in this case THanks !