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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

- Thread starter Albert
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- Thread starter
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- Jan 25, 2013

- 1,225

- Thread starter
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- Jan 25, 2013

- 1,225

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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