Show that H,P,O are collinear where H is the orthocentre CPD and O is the circumcentre of APB The original question is :
The diagonals AC & BD of a cyclic quadrilateral ABCD intersect at P . Let O be the circumcircle of triangle APB and H be the orthocentre of triangle CPD . Show that the points H,P,O are collinear .
I tried this problem in many ways. If we extend PH to meet CD then we get many right angled triangle also extend PO then we get many right angled triangle . Also angle PDC= angle BAP. To shore H,P,O are collinear we shall show that angle HPD= angle BPO. But I can't proceed . Somebody help me .
Thank you
 A: The green dotted lines are perpendicular bisectors to locate O. The blue dotted lines (DD' and PP’) are altitudes to locate H.
From $x = y = \angle APB$, we have the black dotted circle. This further means c = b (= a also).

$\angle 1 = 90^0 - \angle 2 = 90^0 - \angle 2’ = a = b = c$
Result follows because c and $\angle 1$ are vertically opposite.
A: Maybe a bit shorter solution. 
Since $H$ is the orthocenter of CPD, the line $PH$ is the altitude of triangle $CPD$ and is thus perpendicular to $CD$. Draw the line through point $B$ perpendicular to the diagonal $BD$ and let it intersect the line $PH$ at point $E$. If we denote $\angle\, PCD = \beta$ then $\angle \, BPH = 90^{\circ}-\beta = \angle \, APE$. Since $ABCD$ is cyclic, $\beta = \angle \, PCD = \angle \, ACD = \angle \, ABD.$ Since $EB$ is orthogonal to $DB$ by construction, $\angle \, ABE = 90^{\circ} - \angle \, ABD = 90^{\circ} - \beta$. Therefore $\angle \, APE = \angle \, ABE$ so quadrilateral $EBPA$ is inscribed in a circle, which is the circumcircle of triangle $ABP$. Moreover, $\angle \, EBP = 90^{\circ}$ by construction, so $PE$ is a diameter and it passes through the incenter of triangle $ABP$. Consequently, the line $PH \equiv PE$ passes through the incenter of triangle $ABP$, i.e. $H, P$ and the incenter are collinear.  
