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 Xc to denote the complement of X in S, so that Xc is the set of all elements of S that are not in X.
Note that (Xc)c = X, and thus we can think of the subsets of S as coming in pairs:
each subset X is paired with its complement Xc.
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 Xc.
What remains is to show that given any subset X of S, either X is a member of A or else Xc 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 Xc.
Since Y is not disjoint from any member of A, it follows that the set Xc, which contains Y, cannot be disjoint from any member of A.
We conclude from the hypothesis that Xc must be a member of A, and the proof is complete.