Construction of Similar Triangles

Construction of a Triangle Similar to a Given Triangle with a Given Scale Factor

Similar triangles are triangles that have the same shape but are not necessarily the same size. Their corresponding angles are equal, and the lengths of their corresponding sides are in proportion. The ratio of the lengths of corresponding sides is called the scale factor of similarity.

This construction allows us to create a new triangle that is either an enlargement or a reduction of a given triangle, maintaining the exact same shape. The process utilizes the concept of dividing a line segment in a given ratio (I1) and the properties of parallel lines and similar triangles (specifically the Basic Proportionality Theorem).

Construction Steps

Given: A triangle $\triangle ABC$ and a scale factor $k = \frac{m}{n}$, where $m$ and $n$ are positive integers. We want to construct a triangle $\triangle A'B'C'$ similar to $\triangle ABC$ such that the ratio of corresponding sides $\frac{A'B'}{AB} = \frac{B'C'}{BC} = \frac{C'A'}{CA} = k = \frac{m}{n}$.

A common approach is to construct the new triangle with one vertex common to the original triangle. Let's use vertex $A$ as the common vertex and construct $\triangle AB'C'$ such that $B'$ lies on the ray $AB$ (or its extension) and $C'$ lies on the ray $AC$ (or its extension), and $\triangle AB'C' \sim \triangle ABC$.

Tools: Ruler, Compass, Straightedge.

Goal: Construct $\triangle AB'C' \sim \triangle ABC$ with scale factor $m/n$.

Steps:

1. Draw a Ray: From one of the vertices of the given triangle, say $A$, draw a ray $AX$ that makes an acute angle with the side $AB$. Ensure this ray lies on the opposite side of $AB$ from vertex $C$ to avoid clutter within the triangle, unless constructing a smaller triangle inside.
2. Mark Equal Divisions: The scale factor is $m/n$. Determine the maximum of the two numbers, $m$ and $n$. Let $p = \max(m, n)$. Using a compass, choose any convenient radius. Starting from $A$, mark off $p$ equally spaced points along the ray $AX$. Label these points $A_1, A_2, A_3, \dots, A_p$. So, $AA_1 = A_1A_2 = \dots = A_{p-1}A_p$.
3. Join Based on Denominator: Identify the point on the ray $AX$ corresponding to the denominator of the scale factor, which is $A_n$. Join the point $A_n$ to the vertex $B$ of the original triangle. Draw the line segment $A_nB$.
4. Draw Parallel Line for First Side: Identify the point on the ray $AX$ corresponding to the numerator of the scale factor, which is $A_m$. Construct a line through point $A_m$ that is parallel to the line segment $A_nB$. Use the method for constructing a parallel line through a point (I1), by copying the angle $\angle AA_nB$ at point $A_m$. Draw this parallel line. This parallel line will intersect the ray $AB$ (or its extension) at a point. Label this intersection point as $B'$.
5. Draw Parallel Line for Second Side: Now, construct a line through the point $B'$ that is parallel to the side $BC$ of the original triangle $\triangle ABC$. Use the method for constructing a parallel line through a point (I1), by copying the angle $\angle ABC$ at point $B'$. Draw this parallel line. This parallel line will intersect the ray $AC$ (or its extension) at a point. Label this intersection point as $C'$.
6. Resulting Triangle: The triangle $\triangle AB'C'$ with vertices $A$, $B'$, and $C'$ is the required triangle similar to $\triangle ABC$ with the scale factor $m/n$.

The appearance and location of the new triangle $\triangle AB'C'$ relative to $\triangle ABC$ depend on whether the scale factor $k=m/n$ is greater than 1 or less than 1.

Example

Answer:

Cases where Scale Factor is Greater than One

When the scale factor $k = \frac{m}{n}$ is greater than 1 (meaning $m > n$), the constructed triangle $\triangle AB'C'$ will be an enlargement of the original triangle $\triangle ABC$. The new triangle will be larger, and the vertices $B'$ and $C'$ will lie on the extensions of the original sides $AB$ and $AC$, respectively, beyond points $B$ and $C$.

Construction Notes ($m > n$)

Let the scale factor be $m/n$ where $m > n$. You follow the same general steps as outlined in I1, but pay attention to where the parallel lines intersect the original triangle's sides.

1. Draw Ray and Divisions: Draw triangle $\triangle ABC$. Draw ray $AX$ from $A$ making an acute angle with $AB$. Mark $\max(m,n) = m$ points $A_1, \dots, A_n, \dots, A_m$ on $AX$ such that $AA_1 = \dots = A_{m-1}A_m$.
2. Join Denominator Point: Join $A_n$ to $B$.
3. Draw Parallel through Numerator Point: Draw a line through $A_m$ parallel to $A_nB$. Since $A_m$ is further from $A$ than $A_n$ ($m>n$), this parallel line will intersect the extension of the line segment $AB$ beyond $B$. Label the intersection point $B'$. Thus, $B'$ will lie on the ray $AB$ such that $B$ is between $A$ and $B'$. The length $AB'$ will be $(m/n) \times AB$.
4. Draw Second Parallel Line: Draw a line through $B'$ parallel to $BC$. This line will intersect the extension of the line segment $AC$ beyond $C$. Label the intersection point $C'$. Thus, $C'$ will lie on the ray $AC$ such that $C$ is between $A$ and $C'$. The length $AC'$ will be $(m/n) \times AC$.

The resulting triangle $\triangle AB'C'$ is an enlargement of $\triangle ABC$ and contains $\triangle ABC$ if the original triangle is drawn within $\angle XAB$. The ratio of corresponding sides is $\frac{AB'}{AB} = \frac{AC'}{AC} = \frac{B'C'}{BC} = \frac{m}{n} > 1$.

Example

Answer:

Cases where Scale Factor is Less than One

When the scale factor $k = \frac{m}{n}$ is less than 1 (meaning $m < n$), the constructed triangle $\triangle AB'C'$ will be a reduction of the original triangle $\triangle ABC$. The new triangle will be smaller, and the vertices $B'$ and $C'$ will lie on the original sides $AB$ and $AC$, respectively, between points $A$ and $B$ and between $A$ and $C$.

Construction Notes ($m < n$)

Let the scale factor be $m/n$ where $m < n$. You follow the same general steps as outlined in I1, but the location of $A_m$ relative to $A_n$ determines where the parallel lines intersect.

1. Draw Ray and Divisions: Draw triangle $\triangle ABC$. Draw ray $AX$ from $A$ making an acute angle with $AB$. Mark $\max(m,n) = n$ points $A_1, \dots, A_m, \dots, A_n$ on $AX$ such that $AA_1 = \dots = A_{n-1}A_n$.
2. Join Denominator Point: Join $A_n$ to $B$.
3. Draw Parallel through Numerator Point: Draw a line through $A_m$ parallel to $A_nB$. Since $A_m$ is closer to $A$ than $A_n$ ($mline segment $AB$ between $A$ and $B$. Label the intersection point $B'$. Thus, $B'$ will lie on $AB$. The length $AB'$ will be $(m/n) \times AB$.
4. Draw Second Parallel Line: Draw a line through $B'$ parallel to $BC$. This line will intersect the line segment $AC$ between $A$ and $C$. Label the intersection point $C'$. Thus, $C'$ will lie on $AC$. The length $AC'$ will be $(m/n) \times AC$.

The resulting triangle $\triangle AB'C'$ is a reduction of $\triangle ABC$ and lies inside it. The ratio of corresponding sides is $\frac{AB'}{AB} = \frac{AC'}{AC} = \frac{B'C'}{BC} = \frac{m}{n} < 1$.

Example

Answer:

Justification of Similar Triangle Construction

The justification proves that the constructed triangle $\triangle AB'C'$ (or $\triangle A'B'C'$) is similar to the original triangle $\triangle ABC$ with the required scale factor $\frac{m}{n}$. The proof uses the properties of parallel lines cut by transversals and the Basic Proportionality Theorem (BPT) or the AA similarity criterion.

Proof using Parallel Lines and BPT/Similarity

Given:

• Triangle $\triangle ABC$.
• Ray $AX$ making an acute angle with $AB$.
• Points $A_1, A_2, \dots, A_p$ on ray $AX$ such that $AA_1 = A_1A_2 = \dots = A_{p-1}A_p$, where $p = \max(m, n)$. Let the length of each small segment be $k$.
• Segment $A_nB$ is drawn.
• Line $A_mB'$ is drawn through $A_m$ parallel to $A_nB$, intersecting the line containing $AB$ at $B'$. So, $A_mB' \parallel A_nB$.
• Line $B'C'$ is drawn through $B'$ parallel to $BC$, intersecting the line containing $AC$ at $C'$. So, $B'C' \parallel BC$.

To Prove: $\triangle AB'C' \sim \triangle ABC$ and $\frac{AB'}{AB} = \frac{AC'}{AC} = \frac{B'C'}{BC} = \frac{m}{n}$.

Proof:

Step 1: Proving the ratio of sides on ray AX

By construction (Step 2), we marked points on ray $AX$ such that $AA_1 = A_1A_2 = \dots = A_{p-1}A_p = k$.

• The length $AA_m$ consists of $m$ such segments, so $AA_m = mk$.
• The length $AA_n$ consists of $n$ such segments, so $AA_n = nk$.

Therefore, the ratio $\frac{AA_m}{AA_n} = \frac{mk}{nk} = \frac{m}{n}$ (assuming $n \ne 0$, which is true for a scale factor).

Step 2: Proving proportionality using $A_mB' \parallel A_nB$

Consider $\triangle AA_nB$. We have the line segment $A_mB'$ parallel to the side $A_nB$. This line $A_mB'$ intersects the side $AA_n$ (on ray $AX$) at $A_m$ and the side $AB$ (on ray $AB$) at $B'$.

According to the Basic Proportionality Theorem (BPT), if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points (or their extensions), then the other two sides are divided in the same ratio.

Applying BPT to $\triangle AA_nB$ with $A_mB' \parallel A_nB$:

From Step 1, we know $\frac{AA_m}{AA_n} = \frac{m}{n}$. Therefore,

This shows that the ratio of side $AB'$ to $AB$ is equal to the scale factor $m/n$.

Step 3: Proving similarity using $B'C' \parallel BC$

Now consider $\triangle ABC$ and $\triangle AB'C'$. Vertex $A$ is common to both triangles, so $\angle BAC$ is common, i.e., $\angle BAC = \angle B'AC'$.

We constructed line segment $B'C'$ parallel to side $BC$ of $\triangle ABC$. The line $B'C'$ intersects the sides $AB$ and $AC$ (or their extensions) at $B'$ and $C'$ respectively.

Since $B'C' \parallel BC$, and $AB$ is a transversal, the corresponding angles are equal:

Similarly, since $B'C' \parallel BC$, and $AC$ is a transversal, the corresponding angles are equal: Thus, $\triangle AB'C'$ and $\triangle ABC$ have corresponding angles equal: $\angle BAC = \angle B'AC'$, $\angle AB'C' = \angle ABC$, and $\angle AC'B' = \angle ACB$.

Therefore, by AA Similarity Criterion (Angle-Angle Similarity), $\triangle AB'C' \sim \triangle ABC$.

Step 4: Ratio of Sides for Similar Triangles

Since $\triangle AB'C'$ is similar to $\triangle ABC$, the ratio of their corresponding sides must be equal:

From Step 2, we found that $\frac{AB'}{AB} = \frac{m}{n}$.

Combining these results, we have:

Conclusion: The constructed triangle $\triangle AB'C'$ is similar to the given triangle $\triangle ABC$, and the ratio of their corresponding sides is equal to the given scale factor $\frac{m}{n}$. Q.E.D.