IGO 2018 I2

In convex quadrilateral $ABCD$ , the diagonals $AC$ and $BD$ meet at the point $P$ . We know that $\angle DAC = 90^o$ and $2 \angle ADB = \angle ACB$ . If we have $\angle DBC + 2 \angle ADC = 180^o$ prove that $2AP = BP$ .

Proposed by Iman Maghsoudi