Ninth Grade Mathematics: Triangles: Given Base, Sum/Difference of Sides
Constructing a triangle when all three sides (SSS) or two sides and an angle (SAS) are given is straightforward. However, when the data specifies the base, a base angle, and only the sum or difference of the remaining two sides, we must use properties of perpendicular bisectors and isosceles triangles to locate the third vertex. In this chapter, you will master these two classical construction protocols.
Construction 1: Given Base, Base Angle, and Sum of Other Two Sides
We are required to construct a triangle ABC given:
1. Base BC.
2. Base Angle B.
3. The sum of the remaining two sides: AB + AC = s.
Construction Algorithm (Given Base, Base Angle, and Sum of Sides):
1. Draw the base line segment BC of given length.
2. At vertex B, construct the given base Angle B using ray BX.
3. From ray BX, cut off a line segment BD equal to the given sum: BD = AB + AC = s.
4. Join point C to point D (draw segment CD).
5. Construct the perpendicular bisector of line segment CD.
6. Let this perpendicular bisector intersect segment BD at point A.
7. Join point A to point C to complete Triangle ABC!
D (BD = AB + AC)
*
/ \
/ \
/ \
A *-------* (Perpendicular Bisector)
/ \ /
/ \ /
/ \ /
*-------*
B C
AB + AC == BD, AC == AD
Proof of the Construction
Why does this construction produce the exact required triangle?
Proof:
Step 1: Point A lies on the perpendicular bisector of segment CD.
Step 2: By the Perpendicular Bisector Theorem, any point on the perpendicular
bisector of a segment is equidistant from its endpoints:
AD = AC.
Step 3: From the drawing, segment BD is partitioned at point A:
BD = AB + AD.
Step 4: Substitute AC in place of AD:
BD = AB + AC.
Step 5: Since BD was originally constructed to equal the given sum s,
it follows that AB + AC = s.
Triangle ABC has the exact base BC, the exact base Angle B, and the sum of its two sides AB + AC equals the required length.
Construction 2: Given Base, Base Angle, and Difference of Other Two Sides
When the difference of the two sides is given, two distinct geometric cases arise depending on which side is longer: Case 1 where AB > AC (AB - AC = d), and Case 2 where AC > AB (AC - AB = d).
Case 1: When AB > AC (Difference is AB - AC = d)
Algorithm for Case 1 (AB > AC):
1. Draw base BC and construct the given Angle B on ray BX.
2. From ray BX, cut off segment BD equal to the difference: BD = AB - AC = d.
3. Join CD.
4. Construct the perpendicular bisector of segment CD.
5. Let the perpendicular bisector intersect ray BX at point A.
6. Join AC to form Triangle ABC.
Proof of Case 1:
Point A lies on the perpendicular bisector of CD, so AD = AC.
From the ray: BD = AB - AD = AB - AC = d.
The triangle satisfies AB - AC = d.
Case 2: When AC > AB (Difference is AC - AB = d)
Algorithm for Case 2 (AC > AB):
1. Draw base BC and construct given Angle B on ray BX.
2. EXTEND ray BX in the OPPOSITE DIRECTION past vertex B to form a line.
3. From this opposite extended ray, cut off segment BD equal to the difference:
BD = AC - AB = d.
4. Join CD.
5. Construct the perpendicular bisector of CD, intersecting the original ray BX at point A.
6. Join AC to form Triangle ABC.
Proof of Case 2:
Point A lies on the perpendicular bisector of CD, so AD = AC.
From the line: BD = AD - AB = AC - AB = d.
The triangle satisfies AC - AB = d.
Chapter Exercises
Exercise 1: Construct a triangle ABC in which base BC = 7 cm, base Angle B = 75 degrees, and the sum of sides AB + AC = 13 cm. Describe the exact step-by-step construction protocol.
Exercise 2: Construct a triangle ABC in which base BC = 8 cm, Angle B = 45 degrees, and AB - AC = 3.5 cm. Explain why point A must lie on the perpendicular bisector of segment CD.
Exercise 3: Explain the essential difference in procedure between constructing a triangle when AB - AC is given versus when AC - AB is given.
Solutions and Explanations
Solution 1: To construct triangle ABC: Step 1: Draw a straight line segment BC of length 7 cm using a straightedge. Step 2: At vertex B, construct an angle of 75 degrees using compass and straightedge (by bisecting the 30-degree angle between 60 and 90 degrees), extending ray BX. Step 3: With center B and compass radius 13 cm, cut off a point D on ray BX such that BD = 13 cm. Step 4: Draw line segment CD connecting points C and D. Step 5: Construct the perpendicular bisector of line segment CD by drawing arcs of equal radius from C and D that intersect above and below CD, and drawing the line through these intersections. Step 6: Mark point A where the perpendicular bisector intersects segment BD. Step 7: Join point A to point C. Triangle ABC is the required triangle.
Solution 2: Step 1: Draw base BC = 8 cm. Step 2: At vertex B, construct a 45-degree angle (by bisecting a 90-degree angle) on ray BX. Step 3: From ray BX, mark point D such that BD = 3.5 cm. Step 4: Join CD. Step 5: Construct the perpendicular bisector of segment CD. The point of intersection of this perpendicular bisector with ray BX is vertex A. Step 6: Join AC. Point A must lie on the perpendicular bisector of CD because the Perpendicular Bisector Theorem guarantees that any point on the perpendicular bisector is equidistant from the segment's endpoints: AD = AC. Because BD = AB - AD by segment subtraction, substituting AC for AD gives BD = AB - AC = 3.5 cm, ensuring the required side difference.
Solution 3: The essential difference lies in the direction along which the difference segment BD is measured from vertex B. When AB > AC, vertex A lies further along ray BX than point D, so BD = AB - AC is measured directly along ray BX in the upward direction of the angle. When AC > AB, side AC is longer than AB, which means AD = AC must exceed AB. To achieve this, ray BX must be extended in the opposite direction downward past vertex B. Segment BD = AC - AB is measured along this opposite downward extension. Joining CD and erecting its perpendicular bisector intersects the original upward ray BX at vertex A, satisfying BD = AD - AB = AC - AB.