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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 

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.


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
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry
High School Triangles and Other Polygons

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