Prove that under an affine transformation the ratio of…

Prove that under an affine transformation the ratio of segments on two parallel lines is preserved: if $\ell_1\parallel\ell_2$, $A,B\in\ell_1$, $C,D\in\ell_2$ ($C\ne D$) and $\overrightarrow{AB}=\lambda\,\overrightarrow{CD}$, then $\overrightarrow{A'B'}=\lambda\,\overrightarrow{C'D'}$.

Step 1: Images of vectors

Write the map as $\vec r\,'=M\vec r+\vec b$ with $\det M\ne0$. For any points $P,Q$: $\overrightarrow{P'Q'}=(M\vec q+\vec b)-(M\vec p+\vec b)=M\,\overrightarrow{PQ}$.

The translation cancels: vectors are transformed by the linear part alone.

Step 2: Parallel segments

Let $\vec d$ be a common direction vector of $\ell_1$ and $\ell_2$. Then $\overrightarrow{AB}=\alpha\vec d$ and $\overrightarrow{CD}=\beta\vec d$ with $\beta\ne0$, so $\lambda=\alpha/\beta$. By linearity:

\[ \overrightarrow{A'B'}=\alpha\,M\vec d,\qquad \overrightarrow{C'D'}=\beta\,M\vec d . \]

Since $\det M\ne0$, $M\vec d\ne\vec0$. Both image segments are parallel to $M\vec d$, so $\ell_1'\parallel\ell_2'$, and

\[\overrightarrow{A'B'}=\frac{\alpha}{\beta}\,\overrightarrow{C'D'}=\lambda\,\overrightarrow{C'D'}\qquad\blacksquare\]

Step 3: Consequences

• For segments on the same line, $\ell_1=\ell_2$, the simple ratio of three collinear points is preserved. In particular midpoints go to midpoints.
• Lengths themselves are not preserved: $|A'B'|=|\alpha|\,|M\vec d|$ depends on the direction $\vec d$. Only ratios along a single direction are invariant.
• The proof uses only linearity, so it holds unchanged in space (and in $\mathbb R^n$).