Geometry path · G05

Before this lesson: Triangle congruence, Triangle area ratios

Your goal: Prove a midpoint or a parallel segment from the appropriate direction of the theorem.

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.

The midpoint theorem: the key idea

In a triangle, the segment joining the midpoints of two sides is parallel to the third side and half its length. Conversely, a line through the midpoint of one side and parallel to another side bisects the remaining side. A midpoint construction creates both equal lengths and a parallel line, often revealing a hidden parallelogram or a smaller similar triangle. State which two sides contain the midpoints before using the theorem.

Triangle ABC with marked points on its sides.ABCMN
M and N are the side midpoints. MN is parallel to BC and half its length.

A worked example

In triangle ABC, M and N are midpoints of AB and AC. If BC=14, find MN.

The midpoint theorem gives MN∥BC and MN=BC/2=7.

Your turn: change one thing

The three midpoints form a medial triangle. How does its area compare to ABC?

Try this on paper before opening the explanation.

Compare your reasoning

Each side is half the corresponding original side, so the similarity scale is 1/2 and the area scale is (1/2)²=1/4.

Pause and check

A trap to avoid: Applying the converse without both required conditions.

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

Prove the midpoint theorem using similarity.

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

Since AM/AB=AN/AC=1/2 and the included angle at A is common, triangles AMN and ABC are similar by SAS similarity. Corresponding angles are equal, so MN∥BC, and corresponding lengths give MN/BC=1/2.

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. The segment joining two side midpoints is parallel to what?

Foundation

  1. The angle bisector
  2. The third side
  3. Both sides containing the midpoints
  4. Every altitude
Hint

Name the side opposite the common vertex.

Answer and reasoning

The third side. The midpoint theorem makes the connecting segment parallel to the remaining side.

2. Its length is what fraction of the third side?

Foundation

  1. 1/3
  2. 2
  3. 1/4
  4. 1/2
Hint

Use the similarity scale.

Answer and reasoning

1/2. Both adjacent sides have been scaled by 1/2.

3. If BC=14, the corresponding midpoint segment has length what?

Core

  1. 7
  2. 14
  3. 28
  4. 3.5
Hint

Halve the side.

Answer and reasoning

7. MN=14/2=7.

4. What is the medial triangle’s area ratio to the original?

Core

  1. 3/4
  2. 1/4
  3. 1/2
  4. 1/8
Hint

Square the length scale.

Answer and reasoning

1/4. (1/2)²=1/4.

5. Which similarity test proves the midpoint theorem directly?

Stretch

  1. SSA congruence
  2. Only SSS congruence
  3. AAA congruence
  4. SAS similarity
Hint

Two proportional sides enclose the common angle.

Answer and reasoning

SAS similarity. AM/AB=AN/AC and the angle at A is common.

6. A parallel through a side midpoint meets another side where?

Stretch

  1. At the opposite vertex always
  2. At its midpoint
  3. At a trisection point
  4. At any point
Hint

Use the converse midpoint theorem.

Answer and reasoning

At its midpoint. The parallel gives similar triangles with scale 1/2.

7. A midpoint divides a segment in what ratio?

Foundation

  1. 2:3
  2. 1:3
  3. 1:1
  4. 1:2
Hint

The two subsegments are equal.

Answer and reasoning

1:1. Each is half of the whole segment.

8. A midpoint segment has length 9. The parallel third side has length what?

Core

  1. 18
  2. 9
  3. 4.5
  4. 27
Hint

The midpoint segment is half the side.

Answer and reasoning

18. The third side is twice 9.

9. Why is the converse midpoint theorem useful?

Stretch

  1. It assumes every point is a midpoint
  2. It proves circles are equal
  3. It removes the need for a triangle
  4. It proves a new point is a midpoint from parallelism
Hint

One midpoint plus a parallel line fixes the other division ratio.

Answer and reasoning

It proves a new point is a midpoint from parallelism. Similarity forces the second side to be cut in the same 1:1 ratio.

Choose your next step

Continue to Parallel lines and angle bisectors. If this felt difficult, return to a prerequisite above. Every lesson stays open.

Open my revision list →

Original teaching material · IMOolympiad.com. Send a specific correction through our contact page. Learning progress is optional and stays in this browser.