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**thereddevils** 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 $\displaystyle \angle CRD = 90^o$ .

I am not really sure bout this part. I cant see that which triangle is congruent to which in this case

THanks !