MHB Prove Parallelogram Point P's Angles are Equal

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In a parallelogram ABCD with an interior point P, it is given that angles PAB and PCB are equal. By constructing lines P'C parallel to PB and P'P parallel to BC, two new parallelograms, BCP'P and APP'D, are formed. It is established that angles P'DC, PAB, and PCB are equal, leading to the conclusion that points C, P'D, and P are cyclic. Consequently, the angles DPP' and DCP' are shown to be equal, which implies that angles PBA and PDA are also equal. Thus, it is proven that angle PBA equals angle PDA.
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View attachment 972
Point P is an iner point of a parallelogram ABCD
given $\angle PAB=\angle PCB$
please prove :$\angle PBA=\angle PDA$
 

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View attachment 986
Construct P'C//PB , and P'P//BC ,connecting P'D
now
BCP'P and APP'D are two parallelograms
it is easy to see
$\angle P'DC=\angle PAB=\angle PCB=\angle P'PC$
four points C,P'D,P are cyclic
$\therefore \angle DPP'=\angle DCP'$
but $\angle PDA=\angle DPP' , and\,\, \angle DCP'=\angle PBA \therefore \angle PBA=\angle PDA$
 

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