Here is a famous proof, first published by W. W. Rouse Ball in
1892. I will show that every triangle is isosceles. Return to my Mathematics pages
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© Copyright 2002, Jim Loy
Here is a famous proof, first published by W. W. Rouse Ball in
1892. I will show that every triangle is isosceles.
Proof: Draw an arbitrary triangle ABC with AC>BC. I plan to prove that AC=BC. Bisect the angle C, and draw the perpendicular bisector of side AB (bisecting AB at D). These two lines (the bisector of angle C, and the perpendicular bisector of CD) cannot be the same line or be parallel, as then the triangle ABC would be isosceles. So they must meet at some point E. This point E must be either inside the triangle, or outside the triangle, or on the line segment AB. I have drawn all three possible cases, on the left above. In the first two cases, draw the lines AE and BE. In all three cases draw the lines EF and EG, perpendicular to sides AC and BC respectively, as shown in the diagram. Let's examine all three cases.
Case I: Right triangles CEF and CEG are congruent. So, EF=EG and CF=CG. Right triangles ADE and BDE are congruent. So AE=BE. And right triangles AEF and BEG are congruent. So AF=BG. And AC=BC by adding of equals to equals. In other words, triangle ABC is isosceles.
Case II: The same exact argument (except that at the end we subtract equals from equals) shows that this too is isosceles.
Case III: This one is slightly easier, as there are fewer triangles. But the argument is similar.
Triangle ABC is isosceles, in all three cases.
The flaw in the proof: On the right is a more accurate
diagram. By the way, a person can reason out what is wrong with the proof,
without an accurate diagram. But this diagram gives a good clue what is wrong.
You should have been somewhat suspicious that maybe we didn't list all of the
possible cases. Case II has a sub-case. What happens if F is between A and C?
Well the proof doesn't work. We can't add equals to equals, or subtract equals
from equals. We end up adding and subtracting equals, and there is no rule
(axiom) about that. The proof doesn't work.
You can show that E is always outside the triangle, G is outside the triangle, and that F is always between A and C.