the constant for each side of the triangle (on the topic of similarity)

Question

Let's say that two triangles have two corresponding angles that are congruent. How can we then assume that one of the triangles must be multiplied by the same constant on all of its sides to map it to the other triangle?

How do we know for a fact that the constant is the same for each side of the first triangle to map it to the second triangle?

For example:

Let's say that ΔABC ~ ΔDEF by AA similarity.

Which would indicate that DE/AB = EF/BC = DF/AC = K , where K is a constant.

My question is, how do we know that the quotient of dividing corresponding line segments are the same? is there any proof on this?

Answer 1

This is essentially Book VI Proposition 4 in Euclid:

In equiangular triangles the sides about the equal angles are proportional where the corresponding sides are opposite the equal angles.

The image there is

Note how parallel lines come into the proof. That's important, because in noneuclidean geometries there are no parallel lines, or they are not unique, and similar triangles are in fact congruent. You may have to follow some links to see exactly why it answers your question.

Answer 2

suppose $\triangle{ABC}$~$\triangle{DEF}$ which implies $$\angle A = \angle D$$ $$\angle B = \angle E$$ $$\angle C = \angle F$$ by sine rule we know that $$\frac{BC}{\sin{A}}=\frac{AC}{\sin B}=\frac{AB}{\sin C}=K \dots(1)$$ and $$\frac{EF}{\sin D}=\frac{DF}{\sin E}=\frac{DE}{\sin F}=K'$$ or $$\frac{EF}{\sin A}=\frac{DF}{\sin B}=\frac{DE}{\sin C}=K' \dots(2)$$

and by (1)/(2) we get

$$\frac{BC}{EF}=\frac{AC}{DF}=\frac{AB}{DE}=\frac{K}{K'}$$