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Triangle paradox

Here is a nice little 'proof' that all triangles are isosceles triangles......

(1) side BD = DC (construction)
angle BDG = CDG (right angles)
side GD = GD (common)
triangle BDG=CDG (SAS) .. (1)
(2) angle FAG = EAG (construction)
angle AFG = AEG (right angles)
angle AGF = AGE (sum of angles)
triangle AFG = AEG (ASA) .. (2)
(3) side BG = GC (corr. sides from 1)
side GF = GE (corr. sides from 2)
angle GFB = GEC (right angles)
triangle GFB=GEC (RHS) .. (3)
(4) AF = AE (corr. sides from 2)
FB = FC (corr. sides from 3)
AB = AF + FB = AE + FC = AC
ABC is isosceles!
Therefore all triangles are isosceles.
That's interesting. The way you've constructed it, point G is the orthocenter, right? All the altitudes meet there. But isn't it an assumption that the altitude down to BC bisects BC, something that the median should do? Since the altitude is not the same as the median in general, the triangle you constructed is not arbitrary.

So you haven't proved that every triangle is isosceles, you've proved that a triangle whose altitude is the same as the median for one side is in fact isosceles, nevertheless, an interesting result.

I could be wrong about this, my apologies if I am.
Well spotted and concisely described....exactly correct.
I don't know much geometry But
if angle FAG = EAG (construction)
angle AFG = AEG (right angles)

Then they may not always meet at a single point right?
Nice one! AVG spotted it very well. I was a utterly confused after reading the proof for the first time. Smile
As I see it, this whole problem has nothing to do with altitudes (none are constructed) so I don't understand how the orthocenter plays a role here. Point G is just the intersection between the perpendicular bisector of BC and the bisector of BAC, so it's not the orthocenter.

The whole proof actually holds except for one subtle mistake: G is not inside the triangle, except when it is an isosceles. Actually this "proof" proves exactly that (assume G is inside the triangle -> this proof holds -> ABC is an isosceles triangle).

A more constructive proof of this fact can be found by realizing point G must be on the circumscribed circle of ABC:
1. Construct the circumscribed circle of ABC and the perpendicular bisector of BC. Call their intersection H.
2. BH=CH as H is on the perpendicular bisector.
3. A, B, C and H are all on the same circle by construction. From (2) follows that angles BAH and CAH must be equal (property of chords). So H is on the bisector of BAC.
Conclusion: H is the (or just "an" in case of an isosceles) intersection of the perpendicular bisector of BC and the bisector of BAC. It lies on the circumscribed circle and thus outside of the triangle.
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