Can Cecil solve this geometry stumper?
The following geometry problem has haunted me for 30 years. Every math teacher, Ph.D., engineer, etc., I've shown it to was stumped. Now it's your turn. Given any triangle ABC, extend the baseline and bisect the interior and exterior angles at A and B. Extend these four new lines until they intersect at D and E. Prove DCE is a straight line. (It looks straight, but no one can prove it.) How about it?
We ordinarily don't truck with pure (as opposed to applied) mathematics in this column, but since Cecil's stock in the nation's math departments has pretty much gone into the tank after the recent Monty Hall debacle, the time has come to show a little inductive flash. (Sure, it's the testosterone talking, but you ought to see the abuse I got.) This one was made a easier by a handy geometrical fact that, being a master of suspense, I will conceal until later. First the proof:
(1) Extend AC and BC. Put W on BC as shown. Draw segments EX, EY, and EZ perpendicular to AC, CB, and AB.
(2) Triangles EBY and EBZ are congruent (i.e., equal) because of the angle-angle-side theorem. Proof: angle EYB = EZB (both 90), angle EBY = EBZ (given), segment BE = BE.
(3) Therefore segments EY and EZ are congruent.
(4) Triangles AEX and AEZ are congruent. Proof: same reasoning as step 2.
(5) Therefore segments EZ and EX are congruent.
(6) Since EY = EZ and EZ = EX, EY = EX.
(7) Triangles ECX and ECY are congruent because of the hypotenuse-leg theorem. Proof: they are both right triangles; hypotenuse CE = CE; EY = EX (step 6).
(8) Therefore angle XCE = YCE.
(9) Angle WCD = ACD. Proof: proceed as in steps 1-8.
(10) Angle ACD + DCW + WCX = angle ACX = 180 (whole is sum of parts; angle ACX lies on a straight line).
(11) Angle WCA = XCB (vertical angles).
(12) Angle XCE = DCA (both are half of equal angles).
(13) Angle DCW + WCX + XCE = 180 (substitution).
(14) Angle DCE = 180 (whole is sum of parts). QED.
You're now thinking one of two things: (1) who cares? or (2) damn, Cecil sure is smart. Well, don't be too impressed. I knew the feet of the altitudes of any triangle DEF form a triangle ABC whose angles the altitudes bisect. What's more, circles centered on D, E, and F can be drawn such that each is tangent to the lines containing the sides of ABC. (Another tritangent circle can be centered on the point where the altitudes of DEF cross.) Given that, the rest is obvious.
Since this column was originally published I've gotten a half dozen letters from people who think the preceding is inelegant (!) and offer what they regard as a superior alternative. While the proofs vary in the details, they have one common feature: at some point in the proceedings they talk about "triangle ADE" and/or "triangle BDE." In the context of the proof, the implicit assumption is that points D, C, and E lie on one side of this triangle--that they are collinear, in other words. Sorry, folks, but that's what you're supposed to prove.