Let D be the Orthogonal projection of the vertex A of a given triangle.If it stands that [AB]+[BD] ≅ [AC]+[CD] prove that the triangle is equlaterial.
I have tried to prove in the following way but I’m not sure if this is enough: Let it be [AB]+[BD] ≅ [AC]+[CD]....(1) I have supposed that the the triangle is not equlaterial for example let it be scalene traingle Then [AB] ≢ [AC] and [BD] ≢ [CD] so if we add two of give relactions we get that [AB]+[BD] ≢ [AC]+[CD], which is not possible because it contradicts (1) Therefore the triangle is equlaterial. Thank you in advance.
If $a \ne b$ and $c \ne d$, it is still possible that $a+c = b+d$. Take $(a,b,c,d) = (1,2,2,1)$, for example.
Now your proposition is true not only for equilateral triangles, but for any isosceles triangle with $AB=AC$. The proof is a slight modification of yours, by using inequalities instead of unequal signs:
Suppose $AB > AC$. Then by Pythagoreas' Theorem and the fact that $BC \perp AD$, we have:
$$BD = \sqrt {AB^2 - AD^2} > \sqrt{AC^2 - AD^2} = CD$$
Hence $AB + BD > AC + CD$.
If $AB < AC$, $AB + BD < AC + CD$ follows from a similar argument.
Therefore if $AB +BD = AC + CD$, it must be the case that $AB = AC$.
When $AB = AC$, $\triangle ABD \cong \triangle ACD$ by RHS, and thus $AB + BD = AC + CD$, so the implication goes both ways, and we have shown that this proposition is true for any isosceles triangle with $AB = AC$.