Question

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

Answer 1

Proof:

Now, since is perpendicular to and is perpendicular to , we have that the point must lie on the circumcircle of .

We can employ similar arguments to show that lies on the circumcircle of as well as the circumcircle of .

From this we have that , , and are all cyclic quadrilaterals.

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

which in turn tells us that

But is also a cyclic quadrilateral, and so we have

Now, opposite angles in a cyclic quadrilateral are supplementary and so it follows that

Next we need only subtract and we have

.

Recall that both and are cyclic quadrilaterals, and so

and

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

Therefore, , , are collinerar. Q.E.D.

Furthermore, the converse is also true. That is, if the feet of the perpendiculars dropped from a point to the sides of the triangle are collinear, then is on the circumcircle.

Proof: Left to reader as an exercise!

Answer 2

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.