অঙ্কন: একটি রেখাংশকে n টি সমান ভাগে ভাগ করা

Given a segment, we can divide it into $n$ equal parts, where $n$ is any positive integer.

Suppose the segment is the segment $A_0A_n$. We would like to divide it into $n$ congruent segments: $A_0A_1, A_1A_2, ..., A_{n-1}A_n$.

Let $B_0=A_0$. Draw another line through $B_0$. Make a segment of your choice $B_0B_1$.

Copy this segment until you have a series of distinct points $B_0, B_1, B_2, ..., B_n$ such that $B_0B_1=B_1B_2=...=B_{n-1}B_n$.

Now, draw the segment $B_nA_n$. Construct parallel segments to $B_nA_n$ through all of the points $B_1,B_2,...,B_{n-1}$.

Let these parallels intersect $A_0A_n$ in points $A_1,A_2,...,A_{n-1}$. These are our desired points, and the segment has now been divided into $n$ equal segments.

To prove this, see that $B_0B_k=k\cdot B_0B_1$ for all integer $k$ from $1$ to $n$.

Also, due to the parallel lines, $\triangle A_0A_1B_1\sim\triangle A_0A_kB_k$ for all integer $k$ from $1$ to $n$.

Therefore, since the ratios of corresponding sides are equal for similar triangles, we have $A_0A_k=k\cdot A_0A_1$ for all integer $k$ from $1$ to $n$. Remember that $A_0=B_0$!

Thus, we clearly see that $A_0A_1=A_1A_2=...=A_{n-1}A_n$.

We therefore see that this construction solves our problem.