In triangle $ABC$ prove that $AB = 2BC$ In solving this proof I am not permitted to use any numerically related given (i.e. the sum of all angles in a triangle is $180^\circ$ or in a right triangle $a^2 + b^2= c^2$.
Given $\triangle ABC$ is a right triangle with angle $C = 90^\circ$,
Prove that angle $B$ is $2$ times angle $A$, then line segment $AB=2BC$.
 A: Hint Reflect the triangle in side $AC$. Call the reflection of $B$ as $B'$.
Show that if either condition happens, then the triangle $ABB'$ is equilateral. 
[Actually each condition is equivalent to $ABB'$ is equilateral].
A: Let the bisector of $\angle B$ meet side $\overline{AC}$ and $D$. Drop a perpendicular from $D$ to $E$ on $\overline{AB}$.
What can you say about the triangles created?
Edit. Full solution as requested.


*

*Since $\triangle BDC$ and $\triangle BDE$ share a side ($\overline{BD}$), have congruent angles at $B$, and also have congruent right angles, the two triangles themselves are congruent by Side-Angle-Angle. Thus, $\overline{BC}\cong\overline{BE}$.

*Since $\triangle BDE$ and $\triangle ADE$ share a side ($\overline{DE}$), have congruent right angles adjacent to those sides, and also have congruent angles opposite those sides, the triangles themselves are congruent by Side-Angle-Angle. Thus, $\overline{BE}\cong\overline{AE}$.

*So, $E$ separates $\overline{AB}$ into two segments congruent to $\overline{BC}$. Therefore, $|\overline{AB}| = 2|\overline{BC}|$.

