Geometry path · G14

Before this lesson: Triangle congruence, Similar triangles, Circles, tangents and power of a point

Your goal: Give construction steps and prove they produce exactly the required possibilities.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Names you may know: straightedge-and-compass construction.

Triangle Construction with Ruler and Compass: the key idea

A ruler-and-compass construction uses an unmarked straightedge for lines and a compass for circles and transferring lengths. Treat the desired vertex as the intersection of loci: fixed distance from a point gives a circle; fixed distance from a line gives parallel lines; equal distances from two points give a perpendicular bisector. A complete solution has analysis, construction, proof and discussion of existence and multiplicity. A drawing suggests the construction but does not prove it always exists or is unique. Counts should specify whether mirror images across the given base are considered the same.

Two circles intersect above and below a fixed base AB.ABCC′
Equal-radius circles illustrate two mirror-image vertex positions. In an SSS construction use the two radii specified by the side lengths.

A worked example

Construct a triangle with AB=6, AC=5 and BC=4.

Draw AB of length 6. Draw a circle centred at A of radius 5 and one centred at B of radius 4. Their intersections are the possible vertices C. Since |5−4|<6<5+4, there are two intersections symmetric about AB, giving congruent mirror-image triangles. Each intersection has the required distances by construction.

Your turn: change one thing

What changes if AC=2,BC=3 and AB=6?

Try this on paper before opening the explanation.

Compare your reasoning

The circles do not meet because the distance between centres 6 exceeds the sum of radii 5. The triangle inequality fails, so no triangle exists.

Pause and check

A trap to avoid: Giving a plausible drawing without proof, or forgetting impossible and duplicate cases.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

Justify the standard perpendicular-bisector construction for segment AB.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

Draw equal-radius circles centred at A and B with radius greater than AB/2. Their two intersections X,Y are each equidistant from A and B. Let M be the midpoint of AB. Triangles AMX and BMX are congruent by SSS: AM=BM, AX=BX and MX is shared. Thus the two adjacent angles at M are equal and sum to 180°, so each is 90°. The same argument puts Y on the perpendicular through M. Hence XY is that perpendicular bisector.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. A point at fixed distance r from A lies on what locus?

Foundation

  1. A line through A
  2. A parabola always
  3. The perpendicular bisector of every segment
  4. A circle centred at A
Hint

Use the definition of a circle.

Answer and reasoning

A circle centred at A. All points at distance r from A form that circle.

2. A complete construction solution needs what beyond a drawing?

Foundation

  1. Only a ruler measurement
  2. A numerical guess
  3. Proof and an existence discussion
  4. Only colours
Hint

The diagram alone does not justify all cases.

Answer and reasoning

Proof and an existence discussion. The constructed point must satisfy the data, and the allowed number of solutions must be addressed.

3. For base 6 and other sides 5,4, how many vertices lie off the base line if both sides of it count?

Core

  1. 0
  2. 1
  3. 4
  4. 2
Hint

The two circles cross in two symmetric points.

Answer and reasoning

2. Strict triangle inequalities give two mirror-image positions.

4. Can sides 2,3,6 be constructed as a nondegenerate triangle?

Core

  1. Only above the base
  2. No
  3. Yes, uniquely
  4. Yes, twice
Hint

The two smaller sides total 5.

Answer and reasoning

No. 5<6, so the circles are too far apart to intersect.

5. Points equidistant from A and B form which locus?

Stretch

  1. A circle centred at A
  2. The line AB
  3. The perpendicular bisector of AB
  4. The segment AB only
Hint

Use the equal-distance locus.

Answer and reasoning

The perpendicular bisector of AB. The locus is the entire perpendicular bisector.

6. Why use a compass radius greater than AB/2 for the perpendicular-bisector construction?

Stretch

  1. To make the circles disjoint
  2. To make the base longer
  3. To avoid equal radii
  4. To obtain two distinct circle intersections
Hint

At equality the circles only touch.

Answer and reasoning

To obtain two distinct circle intersections. A larger equal radius gives two intersection points determining the bisector line.

7. An unmarked straightedge constructs what directly?

Foundation

  1. A line through two known points
  2. An arbitrary measured length
  3. A circle of any radius
  4. A numerical angle
Hint

It draws lines without a measurement scale.

Answer and reasoning

A line through two known points. Lengths are transferred with a compass, not read from ruler markings.

8. For circles of radii 5 and 4, what centre distance gives external tangency?

Core

  1. 4
  2. 9
  3. 1
  4. 5
Hint

External tangency occurs at the sum of radii.

Answer and reasoning

9. 5+4=9; the single contact produces a degenerate SSS triangle.

9. Why discuss two mirror-image solutions in SSS construction?

Stretch

  1. They have different side lengths
  2. One is always invalid
  3. They are never congruent
  4. The two circle intersections lie on opposite sides of the base
Hint

The data may not specify a side of the base.

Answer and reasoning

The two circle intersections lie on opposite sides of the base. Both positions satisfy the distances, though they are congruent if reflections are identified.

Choose your next step

Try a written problem in the challenge room. If this felt difficult, return to a prerequisite above. Every lesson stays open.

Open my revision list →

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