Where is the error in this proof that isosceles triangles do not exist?
Given: $θ=γ$, then:
$$\cos \theta=\cos \gamma$$
$$\frac{u\cdot(u+v)}{||u||||u+v||}=\frac{(-v)\cdot(-(u+v))}{||v||||u+v||}$$ $$\frac{u\cdot u+u \cdot v}{||u||}=\frac{u \cdot v+v \cdot v}{||v||}$$ $$||v||||u||^2+||v||(u\cdot v)=||u||(u\cdot v)+||u||||v||^2$$ $$||u||||v||(||u||-||v||)=(u\cdot v)(||u||-||v||)$$ $$||u||||v||=u\cdot v$$
So the vectors are parallel? What have I done? This is wrong.
The following line in your working only holds if $||u|| \neq ||v||$, otherwise you are dividing by $0$ when you perform the simplification:
$$ \big{(}||u||\times||v|| \big{)} \big{(}||u||-||v|| \big{)}=\big{(}u\cdot v\big{)}\big{(}||u||-||v||\big{)}$$ $$ \implies||u|| \times ||v||=u\cdot v$$
Therefore, your working is only valid under the strict assumption that the triangle is not isosceles or equilateral. This type of working can be used to prove anything including nonsensical statements like $1=2$.