If simson line of P || AS prove that PA || BC

Question

Question -

If simson line of P with respect to ABC is parallel to AS , prove that PA || BC ...

My try -

I tried menelaus , similarity , angle chasing ...but none of them works ...I can write more thing which I have got but they are not working ... Any hints ?

Thankyou

Answer 1

In the figure, let PA’ produced cut AS produced at T.

The Simpson’s line creates two circles (the red and the green).

z = x and x = y imply x = z. This further means T is another con-cyclic point of PABC.

Then, $\angle APT = 90^0$. Result follows.