Problem:
Points lie on circle and point lies outside the circle. The given points are such that (i) lines and are tangent to , (ii) are collinear, and (iii) . Prove that bisects .
Solution:
Let be the center of circle and let be the midpoint of . It is clear that bisects if and only if are collinear. Consequently, it suffices to show that
The proof is divided into four parts.
- Triangle is isosceles with . (Note that is an isosceles trapezoid and is midpoint of the base . The fact that triangle is isosceles then follows by the Pythagorean Theorem if nothing more elegant comes to mind.) This fact together with Alternate Interior Angles gives
-
Claim. The circle with diameter contains points , and .
Proof. For each of the cases , it is straightforward to verify that is perpendicular to . For it is true that is a right angle because is tangent to the circle at . The same is true for . For , simply use the fact that if is the midpoint of any given chord, then is perpendicular to the chord. -
Referring to the circle , the Inscribed Angle Theorem gives .
-
Because is tangent to at ,
Results from step 1 yield
establishing 2 and completing the proof.
The problems on this page are the property of the MAA's American Mathematics Competitions