# 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

• This question duplicates this, which has a copy of the diagram. (A, B, C there are as here, but D here is P there.) – Rosie F Feb 5 '18 at 18:55

We try the case $4$ for which the point $D$ is outside the triangle. Let $A’$ be the point at which the bisector of angle $A$ cuts the side $BC$ and let $M$ be the midpoint of $BC$.
Supposing $|\overline {BA'}|\lt|\overline{BM}|$. In this case the point $F$ should be inside of the side $AC$ while $E$ should be outside of $AB$. It is true that $|\overline {AE}|=|\overline{AF}|$ however, because of the relative position of points $E$ and $F$ in the sides $AB$ and $AC$ respectively, $|\overline {AE}|\gt|\overline{AB}|$ while $|\overline {AF}|\lt|\overline{AC}|$.This proves that sides $AB$ and $AC$ have not equal lenght (so any argument proving the contrary is false). It is true also that triangles $\triangle DEB$ and $\triangle DFC$ are congruent, however in the case $4$ at least (perhaps in the four cases I guess) it is not question of just subtraction but of both addition and subtraction.
The flaw in the argument (respect to the case $4$) is to have considered only subtraction for both sides instead of addition for a side and subtraction for the other.
In the figure below we give an easily illustrative case with the Pythagorean Triangle of sides $3,4$ and $5$ for which the bisector of the right angle is simply the diagonal $y=x$. 