Are All Triangles Isosceles?
Date: 09/15/1999 at 13:03:37 From: D. Klotz Subject: Are all triangles isosceles? There is a rather involved (and it is claimed, well-known) proof that shows that all triangles are isosceles (it can be found in Euclidean and non-Euclidean geometries - Marvin Jay Greenberg, bottom of pg. 23) - but unfortunately after studying it, I cannot seem to find the flaw in the argument. Your help would be much appreciated. It claims: Given triangle ABC. Construct the bisector of angle A and the perpendicular bisector of side BC opposite to angle A. Now consider the various cases (there are diagrams given in the book). Case 1: The bisector of angle A and the perpendicular bisector of segment BC are either parallel or identical. In either case, the bisector of angle A is perpendicular to BC and hence, by definition, is an altitude. Therefore the triangle is isosceles (The conclusion follows from the Euclidean theorem that states: if an angle bisector and altitude from the same vertex of a triangle coincide, the triangle is isosceles.) Suppose now that the bisector of angle A and the perpendicular bisector of the side opposite are not parallel and do not coincide. Then they intersect in exactly one point, D, and there are 3 cases to consider: Case 2: The point D is inside the triangle Case 3: The point D is on the triangle Case 4: The point D is outside the triangle For each case, construct DE perpendicular to AB and DF perpendicular to AC, and for cases 2 and 4 join D to B and D to C. In each case the following proof now holds: (I don't have the appropriate symbol for congruence on my keyboard so I'll use '=' to mean congruence.) DE = DF because all points on an angle bisector are equidistant from the sides of the angle DA = DA, and angle DEA and angle DFA are right angles Hence triangle ADE is congruent to triangle ADF by the hypotenuse-leg theorem of Euclidean Geometry. Therefore, we have AE = AF. Now, DB = DC because all points on the perpendicular bisector of a segment are equidistant from the ends of the segment. Also, DE = DF, and angle DEB and angle DFC are right angles. Hence, triangle DEB is congruent to triangle DFC by the hypotenuse-leg theorem, and hence FC = BE. It follows that AB = AC, in cases 2 and 3 by addition, and in case 4 by subtraction. The triangle is therefore isosceles. QED
Date: 09/15/1999 at 14:00:47 From: Doctor Rob Subject: Re: Are all triangles isosceles?? Thanks for writing to Ask Dr. Math, D. This is one of the classic fallacious proofs. The error is rather subtle. First of all, cases 2 and 3 are impossible. Secondly, in case 4, either E is between A and B but F is not between A and C, or vice-versa. Thus you get that AB = AE + EB = AF + FC, but AC = AF - FC or else AB = AE - EB = AF - FC, but AC = AF + FC These cannot be equal unless both FC and EB are zero, which cannot happen since we are not in Case 1. If you draw a diagram carefully, using accurate instruments, you will see that what I say is true. Or, you could prove it. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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