Prove that feet of perpendicular from a point on the circumcircle of a triangle on the sides are collinear

Let ABC be a triangle and X be any point on its circumcircle

Also XP, XQ, XR are perpendicular to BC, AB, and AC respectively.

Now, since XP ⊥ PB and XQ ⊥ BQ, 

⇒ X must lie on the circumcircle of BQP. 

Similarly, X must lie on the circumcircle of CPR and AQR.

From this we have that XQBP, XQAR, XRPC are all cyclic quadrilaterals.

So from the fact that XQBP is a cyclic quadrilateral we have that,

∠QXP = 180º - ∠QBP

which in turn tells us that

∠QXP = 180º - ∠ABC

But AXCB is also a cyclic quadrilateral, and so we have

∠AXC = 180º - ∠ABC

⇒ ∠QXP = ∠AXC 

⇒ ∠QXA = ∠PXC 

Similarly XQAR, XRPC are also cyclic quadrilaterals, and

∠PXC = ∠PRC and ∠QXA = ∠QRA 

Now we can combine this with our previous result and we have that

∠PRC = ∠QRA 

Therefore P, Q, Rare collinear.