Convex quadrilateral ABCD which AD=AB+CD

[Mohssine]

ABCD is a convex quadrilateral which AD=AB+CD, P is the intersection of the two bisector or the two angles BAD and ADC, proof that PA=PC

 

[Deleted member 4993]

ABCD is a convex quadrilateral which AD=AB+CD, P is the intersection of the two bisector or the two angles BAD and ADC, proof that PA=PC

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[Dr.Peterson]

ABCD is a convex quadrilateral which AD=AB+CD, P is the intersection of the two bisector of[?] the two angles BAD and ADC, proof that PA=PC

If this is what you mean, then the claim is false:

Please correct the problem.

 

[Mohssine]

Sorry i mean BP=CP

 

[Dr.Peterson]

You told me privately that it should be BP = CP. Here's the new drawing, and a line (green) that you may find helpful: