How to prove Euclid's Proposition 6 from Book I directly? (as opposed to proof by contradiction)

Question

Proposition I.6 states:

If in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another.

This is the converse of Proposition I.5 which says that angles at the base of an isosceles triangle are equal. In Proposition I.6 Euclid derives a contradiction, namely, that the triangle ACB equals a part of itself, triangle DBC, which contradicts Common Notion V, the whole is greater than the part.

How to prove this proposition directly?

Answer 1

Consider the triangle as two triangles $ABC$ and $ACB$ they have side $BC$ in common and two angles $\hat{B}=\hat{C};\;\hat{C}=\hat{B}$ for the rule S-A-A they are equal. Side opposite to equal angles are equal so $BA=CA$.

QED