Tenth Grade Mathematics: Construction of Triangles Similar to a Given Triangle

Scale modeling is one of the premier practical applications of Euclidean geometry, allowing engineers and architects to enlarge or reduce geometric figures while preserving all interior angular relationships. In tenth-grade mathematics, we construct a triangle similar to a given master triangle according to a prescribed scale factor m / n. The construction methodology branches into two distinct cases: constructing a smaller triangle (scale factor < 1, reduction) and constructing a larger triangle (scale factor > 1, enlargement).

The Principle of the Scale Factor

The scale factor (or similarity ratio) m / n dictates the ratio of the side lengths of the new triangle to the corresponding sides of the original triangle:

Scale Factor: k = m / n
1. Reduction (k < 1, m < n): The new triangle lies INSIDE the original triangle
   (e.g., scale factor 3/4 or 2/3).
2. Enlargement (k > 1, m > n): The new triangle lies OUTSIDE the original triangle,
   requiring the extension of the original sides (e.g., scale factor 5/3 or 4/3).

Rule for Compass Points on Auxiliary Ray:
Let the scale factor be m / n.
Always mark max(m, n) equidistant points on the auxiliary acute ray!
- The denominator n ALWAYS connects to the original vertex!
- The numerator m ALWAYS dictates the parallel line that locates the new vertex!

Case 1: Scale Factor Less Than 1 (Reduction, m < n)

Construct a triangle similar to a given triangle ABC with its sides equal to 3/4 of the corresponding sides of triangle ABC.

Step-by-Step Construction (Scale Factor 3/4):
Step 1: Construct the given triangle ABC according to the given measurements.
Step 2: Draw an acute ray BX below side BC (Angle CBX < 90 degrees).
Step 3: Identify max(3, 4) = 4. Mark 4 equidistant points B1, B2, B3, B4 on ray BX:
        B B1 = B1 B2 = B2 B3 = B3 B4.
Step 4: Denominator is 4: Join B4 to vertex C (draw segment B4 C).
Step 5: Numerator is 3: Through B3, draw a line parallel to B4 C, intersecting side BC at C'.
Step 6: Through C', draw a line parallel to side AC, intersecting side AB at A'.
Conclusion: Triangle A' B C' is the required similar triangle with scale factor 3/4.

ASCII Construction Layout (Reduction 3/4):
                 A
                /              A'+---\--+ C' (A'C' || AC)
              / \                /   \               /     \              B-------+---+ C
            \       \                \ B1    \                 \ B2    \                  \ B3 ---+   \ (B3 C' || B4 C)
                \                            \ B4 -------+
                  X

Proof of Construction for Case 1

In triangle B B4 C, line B3 C' is parallel to B4 C. By the Basic Proportionality Theorem (or AA similarity in triangle B B3 C' and triangle B B4 C):

B C' / BC = B B3 / B B4 = 3 / 4

Now in triangle ABC, line A'C' is parallel to AC: Angle B A' C' = Angle BAC (corresponding angles) Angle A' C' B = Angle ACB (corresponding angles) Angle B is common. By the AAA similarity criterion:

triangle A' B C' ~ triangle ABC

Because corresponding sides are proportional:

A'B / AB = B C' / BC = A'C' / AC = 3 / 4

This proves that triangle A' B C' is similar to triangle ABC with sides scaled by 3/4.

Case 2: Scale Factor Greater Than 1 (Enlargement, m > n)

Construct a triangle similar to a given triangle ABC with its sides equal to 5/3 of the corresponding sides of triangle ABC.

Step-by-Step Construction (Scale Factor 5/3):
Step 1: Construct the original triangle ABC. Extend rays BA and BC beyond A and C.
Step 2: Draw an acute ray BX below BC.
Step 3: Max(5, 3) = 5. Mark 5 equidistant points B1, B2, B3, B4, B5 on BX.
Step 4: Denominator is 3: Join B3 to original vertex C (draw B3 C).
Step 5: Numerator is 5: Through B5, draw a line parallel to B3 C, meeting the extended
        ray BC at C'.
Step 6: Through C', draw a line parallel to AC, meeting the extended ray BA at A'.
Conclusion: Triangle A' B C' is the required enlarged triangle with scale factor 5/3.

Chapter Exercises

Exercise 1: Construct a triangle with sides 4 cm, 5 cm, and 6 cm, and then another triangle whose sides are 2/3 of the corresponding sides of the first triangle. Write out the formal steps of construction.

Exercise 2: Construct an isosceles triangle whose base is 8 cm and altitude 4 cm, and then another triangle whose sides are 1 and 1/2 (3/2) times the corresponding sides of the isosceles triangle.

Exercise 3: Explain why, in the construction of similar triangles, the denominator of the scale factor is always connected to the original vertex of the given triangle, while the numerator determines the parallel line.

Solutions and Explanations

Solution 1: First, construct triangle ABC with base BC = 6 cm, side AB = 4 cm (using an arc of 4 cm from B), and side AC = 5 cm (using an arc of 5 cm from C). The scale factor is 2/3 (which is less than 1, so the new triangle is smaller). Draw ray BX making an acute angle with BC. Since max(2, 3) = 3, mark 3 equidistant points B1, B2, B3 on BX. Join B3 to C. Through point B2, draw a line parallel to B3 C by copying Angle B B3 C; let this line meet BC at C'. Through C', draw a line parallel to AC to meet AB at A'. Triangle A' B C' is the required triangle whose sides are 2/3 of the corresponding sides of triangle ABC.

Solution 2: First, construct the original isosceles triangle ABC: draw base BC = 8 cm. Construct the perpendicular bisector of BC to locate its midpoint D. Along the perpendicular bisector, measure altitude DA = 4 cm. Join AB and AC to complete isosceles triangle ABC (where AB = AC = sqrt(4^2 + 4^2) = sqrt(32) = 4*sqrt(2) cm). The scale factor is 3/2 (enlargement). Extend ray BC beyond C and ray BA beyond A. Draw acute ray BX below BC, and mark max(3, 2) = 5 equidistant points B1, B2, B3 on BX. Denominator is 2: join B2 to vertex C. Numerator is 3: through B3, draw a line parallel to B2 C to intersect extended ray BC at C'. Through C', draw a line parallel to AC to intersect extended ray BA at A'. Triangle A' B C' is the required similar triangle with sides 1.5 times the original.

Solution 3: In the scale factor m / n, the denominator n represents the total number of parts corresponding to the original triangle, while the numerator m represents the number of parts corresponding to the new similar triangle. By the Basic Proportionality Theorem, when a line is drawn from the denominator point Bn to the original vertex C, the segment B Bn represents the full measure of the original base BC. Drawing a parallel line from the numerator point Bm to intersect the line of the base at C' sets up the proportion B C' / BC = B Bm / B Bn = m / n. Thus, connecting the denominator to the original vertex establishes the baseline unit, allowing the numerator point to project the scaled vertex proportionally.