Dividing a Line Segment

Division of a Line Segment in a Given Ratio (Internally)

To divide a line segment internally in a given ratio $m:n$ means to find a point $C$ on the segment $AB$ such that the ratio of the length of the segment $AC$ to the length of the segment $CB$ is equal to the given ratio $m:n$. The point $C$ lies *between* $A$ and $B$. This construction is based on the principle of proportionality arising from parallel lines cut by transversals, specifically related to the Basic Proportionality Theorem (BPT).

Construction Steps

Given: A line segment $AB$ and a ratio $m:n$, where $m$ and $n$ are positive integers.

Tools: Compass, Straightedge.

Goal: Find a point $C$ on the line segment $AB$ such that $AC:CB = m:n$.

Steps:

1. Draw a Ray: Draw any ray $AX$ starting from point $A$ and making an acute angle ($\angle BAX$) with the line segment $AB$. The direction of this ray does not matter beyond making an acute angle.
2. Mark Equal Segments on Ray AX: The given ratio is $m:n$. The total number of equal parts needed is $m+n$. Using a compass, choose any convenient radius (compass width). Starting from point $A$, mark off $m+n$ points on the ray $AX$ such that the distance between consecutive points is equal. Label these points $A_1, A_2, A_3, \dots, A_m, A_{m+1}, \dots, A_{m+n}$. So, $AA_1 = A_1A_2 = A_2A_3 = \dots = A_{m+n-1}A_{m+n}$.
3. Join the Endpoint: Use the straightedge to join the point $A_{m+n}$ (the $(m+n)^{th}$ point on ray $AX$) to the other endpoint of the given line segment, point $B$. Draw the line segment $A_{m+n}B$.
4. Draw a Parallel Line: Locate the $m^{th}$ point on the ray $AX$, which is $A_m$. Construct a line through $A_m$ that is parallel to the line segment $A_{m+n}B$. This parallel line will intersect the original line segment $AB$. Use the method for constructing a parallel line through a point (I1), by copying the angle $\angle AA_{m+n}B$ at the point $A_m$. Draw this line. Label the point where this parallel line intersects the line segment $AB$ as $C$.

The point $C$ obtained on the line segment $AB$ is the required point that divides $AB$ internally in the ratio $m:n$. This means that the ratio of the length of $AC$ to the length of $CB$ is $\frac{AC}{CB} = \frac{m}{n}$.

Example

Answer:

Justification of Line Segment Division Construction

The justification proves that the point $C$ obtained by the construction method described in I1 indeed divides the line segment $AB$ internally in the given ratio $m:n$, i.e., $\frac{AC}{CB} = \frac{m}{n}$. The proof relies on the Basic Proportionality Theorem (BPT), also known as Thales' Theorem.

Proof using Basic Proportionality Theorem (Thales' Theorem)

Given:

• Line segment $AB$.
• Ray $AX$ making an acute angle with $AB$.
• Points $A_1, A_2, \dots, A_m, \dots, A_{m+n}$ on ray $AX$ such that $AA_1 = A_1A_2 = \dots = A_{m+n-1}A_{m+n}$. Let the length of each of these small segments be $k$ units.
• Line segment $A_{m+n}B$ is joined.
• A line is drawn through $A_m$ parallel to $A_{m+n}B$, intersecting $AB$ at $C$. So, $A_mC \parallel A_{m+n}B$.

To Prove: $\frac{AC}{CB} = \frac{m}{n}$.

Proof:

Consider $\triangle ABA_{m+n}$. In this triangle, the line segment $A_mC$ is constructed such that it is parallel to the side $A_{m+n}B$. The line $A_mC$ intersects the side $AB$ at $C$ and the side $AA_{m+n}$ (which is part of the ray $AX$) at $A_m$.

According to the Basic Proportionality Theorem (BPT), if a line parallel to one side of a triangle intersects the other two sides in distinct points, then it divides the two sides proportionally.

Applying the BPT to $\triangle ABA_{m+n}$ with line $A_mC \parallel A_{m+n}B$, we get the following proportion:

Now, let's examine the lengths on the ray $AX$ based on the construction steps (Step 2).

By construction, we marked $m+n$ equally spaced points on $AX$. Let the length of each small segment $AA_1, A_1A_2, \dots$ be $k$.

• The segment $AA_m$ is composed of $m$ such equal segments:

  $\text{Length of } AA_m = AA_1 + A_1A_2 + \dots + A_{m-1}A_m$ ($m$ segments)

  ... (i)

  $\text{Length of } AA_m = m \times k$

  ... (ii)

• The segment $A_m A_{m+n}$ is composed of the segments from $A_m$ to $A_{m+1}$, up to $A_{m+n-1}$ to $A_{m+n}$. There are $(m+n) - m = n$ such equal segments:

  $\text{Length of } A_m A_{m+n} = A_mA_{m+1} + \dots + A_{m+n-1}A_{m+n}$ ($n$ segments)

  ... (iii)

  $\text{Length of } A_m A_{m+n} = n \times k$

  ... (iv)

Substitute the expressions for $AA_m$ and $A_m A_{m+n}$ from equations (ii) and (iv) into the BPT proportion:

Assuming the compass width $k > 0$, we can cancel $k$ from the numerator and denominator:

This shows that the point $C$ divides the line segment $AB$ in the given ratio $m:n$.

Alternative Method (Using a Second Ray):

Sometimes, the construction is done by drawing a second ray $BY$ parallel to $AX$ from point $B$, making an acute angle $\angle ABY$ on the opposite side of $AB$. $m$ equal segments are marked on $AX$ (say $A_1, \dots, A_m$) and $n$ equal segments are marked on $BY$ (say $B_1, \dots, B_n$) using the same compass width. The point $A_m$ is then joined to $B_n$. The intersection of $A_mB_n$ with $AB$ gives the point $C$.

Justification of Alternative Method (Using Similar Triangles):

Consider $\triangle AA_mC$ and $\triangle BB_nC$.

• $AA_m \parallel BB_n$ (By construction of parallel rays or alternate approach where $BY$ is constructed parallel to $AX$)
• $\angle AAC_m = \angle BBC_n$ (Alternate Interior Angles if AX || BY and transversal AB) OR $\angle XAB = \angle YBA$ (Alternate Interior Angles if AX || BY and transversal AB)
• $\angle ACm A_m = \angle BCn B_n$ (Alternate Interior Angles if $A_mB_n$ is the transversal crossing $AA_m$ and $BB_n$ IF $AA_m \parallel BB_n$ which is not generally true in this construction)
• $\angle ACA_m = \angle BCB_n$ (Vertically Opposite Angles if $A_mB_n$ is a straight line and $AB$ is a straight line)
• $\angle CA_mA = \angle CB_nB$ (Alternate Interior Angles, since $A_mB_n$ is a transversal intersecting parallel lines $AA_m$ and $BB_n$ IF $AA_m \parallel BB_n$. However, $AA_m$ and $BB_n$ are typically segments on rays, not necessarily parallel lines themselves).

A simpler justification for the alternative method is to use similar triangles formed by the transversal $A_mB_n$ intersecting the parallel rays $AX$ and $BY$.

Since $AX \parallel BY$, and $A_mB_n$ is a transversal, $\angle A A_m B_n = \angle B B_n A_m$ (Alternate Interior Angles if $A_mB_n$ was a transversal between lines $AX$ and $BY$). Also, $\angle XAB = \angle YBA$ (if rays are constructed parallel). And $\angle ACA_m = \angle BCB_n$ (Vertically Opposite Angles).

Consider $\triangle CA_mA$ and $\triangle CB_nB$.

The angles $\angle CA_m A$ and $\angle CB_n B$ are formed by the transversal $A_mB_n$ cutting the lines $AA_m$ and $BB_n$. If rays $AX$ and $BY$ are parallel, then lines containing $AA_m$ and $BB_n$ are parallel.

Therefore, by AA similarity criterion, $\triangle CA_mA \sim \triangle CB_nB$.

For similar triangles, the ratio of corresponding sides is equal:

By construction, $AA_m = m \times k$ and $BB_n = n \times k$ (using the same compass width $k$ for both rays $AX$ and $BY$).

Substitute these into the ratio:

This confirms the division in the ratio $m:n$ using the alternative method based on similar triangles formed by parallel rays.

Conclusion: Both methods of construction validly divide the line segment in the given ratio $m:n$, justified by the Basic Proportionality Theorem or properties of similar triangles arising from parallel lines.