**Example 1**. In the diagram, *AP* is an altitude of ∆*ABC*, and point *Q* is chosen on side *BC* so that ∠*BAQ* = ∠*CAP*. Show that line *AQ* goes through the circumcentre of ∆*ABC*.

**Example 2**. Let *S *be a finite set and recall that two subsets *X *and *Y *of *S *are said to be *disjoint *if they have no elements in common. Suppose that a collection *A* of subsets of *S *has the property that no two of the sets in *A *are disjoint but that every subset of *S *that is not in *A *is disjoint from some member of *A*. Prove that *A *contains exactly half of the subsets of *S*.

**Solution 1**. Draw the circumcircle and extend *AQ *to meet the circle at *R*. Then draw chord *BR*, as shown in figure.

To establish the given result, we must show that *AR *is a diameter of the circle, and for this it suffices to show that ∠*ABR *= 90°.

Comparing ∆*APC *and ∆*ABR*, we see that

∠

BAR= ∠PAC, by hypothesis;∠

ACP =∠ARB, because these angles subtend the same arc on the circle.

It follows that the third angles of the two triangles are equal.

Therefore, ∠*ABR = *∠*APC *= 90°, as desired.

**Solution 2**. If *X* is a subset of *S*, we write *X ^{c}* to denote the complement of

*X*in

*S*, so that

*X*is the set of all elements of

^{c}*S*that are not in

*X*.

Note that (*X ^{c}*)

*=*

^{c}*X*, and thus we can think of the subsets of

*S*as coming in pairs:

*each subset X is paired with its complement X ^{c}.*

The number of pairs, therefore, is exactly half of the total number of subsets.

To complete the proof, we will show that the collection *A* consists of exactly one set from each such pair.

No two members of *A* are disjoint, and thus *A* cannot contain both a subset *X* and its complement *X ^{c}*.

What remains is to show that given any subset *X* of *S*, either *X* is a member of *A* or else *X ^{c}* is a member of

*A*. If

*X*is not in

*A*, then by assumption, there is a member

*Y*of

*A*disjoint from

*X*, and thus

*Y*is contained in

*X*.

^{c}Since *Y* is not disjoint from any member of *A*, it follows that the set *X ^{c}*, which contains

*Y*, cannot be disjoint from any member of

*A*.

We conclude from the hypothesis that *X ^{c}* must be a member of

*A*, and the proof is complete.