Prove that in a general triangle ABC if two bisectors are equal, this triangle must be isosceles.
(Proposed by Ali Abouzar/FB/Math)
The statement is known as the ”Steiner-Lehmus theorem”. There is much material on it on the web, including proofs. In Coxeter’s book ”Geometry revisited” there is a proof that is elementary as well as easy.
Proof by geometry (my own though certainly many people have constructed it or something similar before. After all the problem is a theorem that has attracted interest for well over a century).
First we derive a formula for the length of a bisector (I got the idea for this from my friend Tomas Carnstam, Lund University, Sweden). – Let a, b, c be the sides of a triangle and L the length of the bisector from C. Then two expressions for the triangle area gives
bL sin(C/2) + aL sin(C/2) = ba sin C.
This gives together with the formula sin(2v) = 2 sin v cos v, L = 2ab/(a + b)·cos(C/2).
Using this formula, if the bisectors from C and B are equally long,
2ab/(a + b)·cos(C/2) = 2ac/(a + c)·cos(B/2). Square this to get
b²(a + c)²·cos²(C/2) = c²(a + b)²·cos²(B/2)
b²(a + c)²·(1 + cos C) = c²(a + b)²·(1 + cos B). Then insert (by the cosine theorem)
cos C = (a² + b² – c²)/(2ab) etc, to get
b²(a + c)²·(1 + (a² + b² – c²)/(2ab)) = c²(a + b)²·(1 + (a² + c² – b²)/(2ac))
b(a + c)²·(2ab + a² + b² – c²) = c(a + b)²·(2ac + a² + c² – b²)
b(a + c)²(a + b – c) = c(a + b)²(a + c – b)
Then it is straightforward to expand and then factorize to get
(b – c)(a³ + b²c + bc² + a²b + a²c + 3abc) = 0.
Here the second factor is obviously positive so b – c = 0, i e b = c.