If two triangles have the same angles they are proportional, therefore the ratios of their sides are constant. However, I do not know how to prove that the ratios must be constant from the fact that their angles are equal.
How would you generalize the proof to other figures?
On
Generalization $n>3$ needs caution. If there are two $n$ sided polygons with same angles in the same order then please note the ratio of their corresponding sides will not, in general be in proportion.
For constant proportion of corresponding sides we need full homothety and similitude.. as is violated in case of the red exaggerated parallel displacement of one side.
It was brought to my attention that my original argument may use circular reasoning when I introduce the sine function. I have prepared a second argument, hopefully more sound than the first.
New argument: Consider the following diagram.
Line $ED\parallel BC$ and it is easy to verify that all of the angles in $\triangle ABC$ are equal to all of the angles in $\triangle AED$.
Since line $ED\parallel BC$, then the slopes describing each line must be equal, that is, $\frac{E_y}{D_x-E_x}=\frac{B_y}{C_x-B_x}$. Using the distance formula, $$\begin{align*}||ED||^2 &= (D_x-E_x)^2+(E_y)^2 \\ &= \bigg(\frac{E_y(C_x-B_x)}{B_y}\bigg)^2 + (E_y)^2\\&= \bigg(\frac{E_y}{B_y}\bigg)^2\big((C_x-B_x)^2 +(B_y)^2\big)\\&=\bigg(\frac{E_y}{B_y}\bigg)^2||BC||^2 \\ &\Longrightarrow ||ED||=\frac{E_y}{B_y}||BC||\end{align*} $$
Note that points $A, E$, and $B$ are co-linear because the line $ED$ intersects line $AB$ at point $E$ and a similar argument follows:$\frac{E_y}{E_x}=\frac{B_y}{B_x}$. Using the distance formula again, $$\begin{align*}||AE||^2 &= (E_y)^2+(E_x)^2\\&=(E_y)^2+\bigg(\frac{B_xE_y}{B_y}\bigg)^2\\&=\bigg(\frac{E_y}{B_y}\bigg)^2((B_y)^2+(B_x)^2)\\&=\bigg(\frac{E_y}{B_y}\bigg)^2||AB||^2\\&\Longrightarrow ||AE||=\frac{E_y}{B_y}||AB|| \end{align*} $$
For the last two legs, notice that we can manipulate the equation $\frac{E_y}{D_x-E_x}=\frac{B_y}{C_x-B_x}$ into $$D_x=\frac{E_y}{B_y}\bigg[C_x+\frac{E_xB_y-B_xE_y}{E_y}\bigg]$$
However, we must realize that the expression $E_xB_y-B_xE_y$ is the determinant of two co-linear points, which is $0$, so $$D_x=\frac{E_y}{B_y}C_x$$ which is also $$||AD||=\frac{E_y}{B_y}||AC||$$ and all of the lengths of corresponding sides are proportional to each other by the same scaling factor.
Original Argument: Suppose we have triangles $\triangle ABC$ and $\triangle A^{\prime}B^{\prime}C^{\prime}$, where the capital letters represent the vertices of the triangles (I have pictured the triangles drawn from vertex A to vertex B to vertex C in a counterclockwise fashion, with A being the left most vertex). Then each triangle has sides $a,b,c$ and $a^{\prime}, b^{\prime}, c^{\prime}$ across from their corresponding vertices.
Let the angle at vertices $A$ and $A^{\prime}$ be $\alpha$, $B$ and $B^{\prime}$ be $\beta$, and $C$ and $C^{\prime}$ be $\gamma$. Without loss of generality, drop an altitude in $\triangle ABC$ from vertex $C$ to side $c$ at point $P$. Let $PB=x$ and $CP=h$. Do the same with $\triangle A^{\prime}B^{\prime}C^{\prime}$. Then $\triangle PCB$ is a right triangle with hypotenuse $a$ and legs $h$ and $x$.
Since the angle formed by sides $c$ and $a$, and $c^{\prime}$ and $a^{\prime}$ and is $\beta$, then $\sin{\beta}$ must be the same for both triangles.
$$\sin{\beta}=\frac{h}{a}=\frac{h^{\prime}}{a^{\prime}}$$
Thus $h^{\prime}=kh$ and $a'=ka$, where $k$ is some scaling factor.
Similarly, $$\sin{\alpha} = \frac{h}{b} = \frac{h'}{b'}$$
Then $b'=sb$ where $s$ is some scaling factor. But if $h'=kh$, then $\frac{kh}{b'}=\frac{h}{b}$ and $b'=kb$. So $s=k$ and the scaling factors are equal. You could do this once more to show that $c'=kc$ by dropping a perpendicular from a different vertex.
We conclude with $$\frac{a'}{a}=\frac{b'}{b}=\frac{c'}{c}=k$$
As another commenter stated, you could decompose a figure into triangles and then and then scale each individual triangle by some scaling factor to achieve a similar scaled figure. You should reference homothecy or homothety. Here is another answer that is closely related to this one.