Geometric Constructions
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective
Construct triangles, perpendicular bisectors, angle bisectors, and standard angles accurately using only a pair of compasses and a straightedge (ruler without measurement markings).
The Fundi's Master Tool: Pure Geometric Symmetry
Imagine a master carpenter (fundi) in Gikomba constructing a heavy wooden roof truss for a school hall. The fundi has no laser level or digital protractor—only a taut string (acting as a compass) and a straight timber plank (straightedge). By striking intersecting arcs from both ends of the tie-beam, the fundi discovers the true vertical centre-line with zero measurement error. This ancient method relies on exact symmetry rather than approximate measuring tape readings.
(a) Concrete Scenario: Finding the True Midpoint
A builder in Kisumu needs to erect a central support pillar at the exact midpoint of a \(10\text{ m}\) foundation beam \(AB\). If they measure with a sagging tape, errors accumulate. Instead, opening a compass wider than half of \(AB\) and striking equal arcs from \(A\) and \(B\) yields two points \(P\) and \(Q\) above and below the beam. Joining \(P\) to \(Q\) cuts \(AB\) at exactly \(90^\circ\) and splits \(AB\) into two perfectly equal halves.
(b) Geometric Principle: The Locus of Equidistant Points
Why does this work? Every point on the perpendicular bisector of a line segment \(AB\) is at the exact same distance from \(A\) as it is from \(B\). When we draw arcs of equal radius \(r\) from \(A\) and \(B\):
- Upper intersection \(P\) satisfies \(PA = PB = r\)
- Lower intersection \(Q\) satisfies \(QA = QB = r\)
- Quadrilateral \(APBQ\) is a rhombus, whose diagonals are perpendicular bisectors of each other.
(c) Algebraic Relationship
For any point \(X\) lying on the perpendicular bisector of line segment \(AB\):
\[XA = XB\]For an angle bisector of \(\angle ABC\), every point \(P\) on the bisector ray is equidistant from the arms \(BA\) and \(BC\):
\[d(P, BA) = d(P, BC)\]Interactive Compass Construction Simulation
Step through the perpendicular bisector construction to see how equal radii create symmetric crossing points.
Click 'Step 1: Line AB' to begin.
Key Formulas
1. Perpendicular Bisector Equidistance
\[PA = PB\]Any point \(P\) on the perpendicular bisector of line segment \(AB\) is equidistant from endpoints \(A\) and \(B\). Diagonals of the rhombus formed by the arc radii intersect at \(90^\circ\) and bisect each other.
2. Internal Angle Bisector Theorem
\[\frac{BD}{DC} = \frac{AB}{AC}\]In \(\triangle ABC\), if the ray from \(A\) bisects \(\angle BAC\) and meets side \(BC\) at point \(D\), it splits the opposite side \(BC\) in the exact ratio of the two adjacent sides \(AB\) and \(AC\).
3. Special Angle Construction Ratios
- \(60^\circ\) Angle: Formed by constructing an equilateral triangle where base \(= r\) and intersecting arc radius \(= r\).
- \(30^\circ\) Angle: Obtained by bisecting a \(60^\circ\) angle.
- \(90^\circ\) Angle: Erected as a perpendicular from a point on a line.
- \(45^\circ\) Angle: Obtained by bisecting a \(90^\circ\) angle.
- \(75^\circ\) Angle: Constructed as \(60^\circ + 15^\circ\) (bisecting the \(30^\circ\) interval between \(60^\circ\) and \(90^\circ\)).
- \(105^\circ\) Angle: Constructed as \(90^\circ + 15^\circ\) or \(180^\circ - 75^\circ\).
4. Inscribed Regular Hexagon
\[\text{Side Length } s = r, \quad \text{Central Angle} = 60^\circ, \quad \text{Interior Angle} = 120^\circ\]Stepping around the circumference of a circle of radius \(r\) using the same compass width \(r\) precisely marks out the 6 vertices of a regular hexagon.
Worked Examples
Example 1 (Easy): Bisecting a Given Angle
Problem: Given an angle \(\angle ABC = 70^\circ\), outline the step-by-step compass construction to bisect it and find the measure of each resulting angle.
- Step 1: Place the sharp compass point at vertex \(B\) and draw an arc of arbitrary radius intersecting ray \(BA\) at \(X\) and ray \(BC\) at \(Y\).
- Step 2: Without changing the compass width (or setting a convenient fixed radius), place the compass at \(X\) and swing an arc inside the interior of \(\angle ABC\).
- Step 3: With the exact same compass width, place the point at \(Y\) and strike a second arc intersecting the first at point \(Z\).
- Step 4: Use a straightedge to draw the ray \(BZ\). Ray \(BZ\) is the angle bisector.
Calculation:
\[\angle ABZ = \angle CBZ = \frac{70^\circ}{2} = 35^\circ\]Example 2 (Medium): Internal Division of a Line Segment
Problem: A line segment \(PQ = 15\text{ cm}\) is to be divided internally into 5 equal parts using a compass and straightedge. Find the distance from \(P\) to the 2nd division point \(K\).
- Step 1 (Auxiliary Ray): Draw a ray \(PR\) making an acute angle (e.g., \(30^\circ\)) with \(PQ\).
- Step 2 (Equal Marks): Using a compass set to a fixed small width, make 5 consecutive equal marks along \(PR\): \(P_1, P_2, P_3, P_4, P_5\).
- Step 3 (Terminal Connection): Join \(P_5\) to \(Q\) using a straightedge.
- Step 4 (Parallel Lines): Using compass angle-copying or set-square sliding, draw lines through \(P_1, P_2, P_3, P_4\) parallel to \(P_5Q\).
- Step 5: The line from \(P_2\) meets \(PQ\) at \(K\), dividing \(PQ\) such that \(PK : KQ = 2 : 3\).
Calculation:
\[PK = \frac{2}{5} \times 15\text{ cm} = 6\text{ cm}\]Example 3 (Hard): Finding Distance Between Arc Crossing Points
Problem: A line segment \(AB = 10\text{ cm}\). With centres \(A\) and \(B\), two circular arcs of radius \(13\text{ cm}\) are drawn, intersecting at points \(C\) above \(AB\) and \(D\) below \(AB\). Calculate the exact length of the vertical segment \(CD\).
- Geometric Reasoning: The line \(CD\) is the perpendicular bisector of \(AB\). Let \(M\) be the intersection of \(AB\) and \(CD\).
- Because \(M\) is the midpoint of \(AB\): \[AM = MB = \frac{10}{2} = 5\text{ cm}\]
- In right-angled triangle \(\triangle AMC\), by Pythagoras' Theorem: \[AC^2 = AM^2 + MC^2\]\[13^2 = 5^2 + MC^2\]\[169 = 25 + MC^2 \implies MC^2 = 144 \implies MC = 12\text{ cm}\]
- By symmetry, \(MD = MC = 12\text{ cm}\). Therefore, the full distance \(CD\) is: \[CD = MC + MD = 12 + 12 = 24\text{ cm}\]
Common Mistakes
Misconception 1: Changing the Compass Radius Mid-Construction
Misconception 2: Assuming Angle Bisector Always Halves the Opposite Side
Misconception 3: Erasure of Construction Arcs
Real World
1. Roof Truss Fabrication in Kenyan Construction
When building timber or light-gauge steel trusses for homes in Nakuru or Eldoret, fabricators use string lines and fixed rods as trammel compasses. To find the exact vertical king-post line without an error-prone tape measure on a \(12\text{ m}\) span, the carpenter strikes equal arcs from both wall plates. The apex joint constructed this way guarantees that the roof load is distributed with perfect symmetry to load-bearing walls.
2. Road Junction Layout and Land Surveying
Civil engineers laying out a curved slip road or dividing an irregular triangular parcel of land in Machakos utilize geometric bisection. Constructing the bisector of an intersecting boundary establishes a neutral drainage line equidistant from adjacent boundaries, preventing soil erosion and boundary disputes.
3. Solar Panel Orientation
To maximize photovoltaic yield, solar installers determine true Solar Noon (geometric South/North). By placing a vertical gnomon (peg) on the ground and marking two points where morning and afternoon shadows have identical lengths, the perpendicular bisector of the segment between the two points gives the exact True Meridian line.
Practice