The Midpoint Theorem — Grade 10 Maths (CAPS)

The line joining the midpoints of two sides of a triangle is parallel to the third side and half its length — the theorem, its converse, and how to use each in a proof.

This is one of the most useful results in Grade 10 geometry, and it is short enough to learn word for word.

1The theorem

D and E are the midpoints of AB and AC. Then DE is parallel to BC and exactly half its length.

If D and E are the midpoints of AB and AC, then
DE ∥ BC and DE = ½BC

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Say it as one sentence: "the line joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it." That exact wording is the reason you quote in a proof.

2Using it to find a length

Worked ExampleD and E are midpoints of AB and AC. BC = 14 cm. Find DE.

1
DE = ½BC (midpoint theorem)
Quote the theorem as your reason.
2
DE = ½ × 14.
3
DE = 7 cm.

Worked ExampleD and E are midpoints of AB and AC. DE = 6,5 cm. Find BC.

1
DE = ½BC, so BC = 2 × DE.
Working backwards, double it.
2
BC = 13 cm.

3The converse

The theorem also works in reverse, and the converse is examined just as often.

If a line through the midpoint of one side is parallel to a second side,
then it bisects the third side.

Worked ExampleD is the midpoint of AB and DE ∥ BC. Prove that E is the midpoint of AC.

1
Statement: D is the midpoint of AB (given)
2
Statement: DE ∥ BC (given)
3
∴ AE = EC, so E is the midpoint of AC (converse of the midpoint theorem)
Name the converse explicitly — that is what earns the reason mark.

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The theorem needs two midpoints; the converse needs one midpoint plus a parallel line. Check which one you have been given before quoting a reason.

4Why it works

Extend DE to F so that DE = EF, and join FC.
In △ADE and △CFE: AE = EC (E is the midpoint of AC), DE = EF (construction) and AÊD = CÊF (vertically opposite angles), so △ADE ≡ △CFE (SAS).
Therefore FC = AD = DB, and AD̂E = CF̂E makes FC ∥ DB. DBCF then has one pair of opposite sides both equal and parallel, so DBCF is a parallelogram.
Hence DF ∥ BC and DF = BC. Since DE = ½DF, it follows that DE ∥ BC and DE = ½BC.

It also follows that △ADE ||| △ABC: DE ∥ BC gives equal corresponding angles and Â is common, so the two triangles are equiangular. Every side of △ADE is half the matching side of △ABC, and its area is a quarter of the area of △ABC — areas scale by the SQUARE of the side ratio.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
State the midpoint theorem.
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The line joining the midpoints of two sides of a triangle is parallel to the third side and half its length.
2
D and E are midpoints of AB and AC; BC = 14 cm. Find DE.
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7 cm
3
D and E are midpoints of AB and AC; BC = 25 cm. Find DE.
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12,5 cm
4
D and E are midpoints; DE = 6,5 cm. Find BC.
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13 cm
5
D and E are midpoints; DE = 9 cm. Find BC.
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18 cm
6
State the converse of the midpoint theorem.
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A line through the midpoint of one side, parallel to a second side, bisects the third side.
7
D is the midpoint of AB and DE ∥ BC. What can you conclude about E?
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E is the midpoint of AC (converse of the midpoint theorem).
8
Why are △ADE and △ABC similar?
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DE ∥ BC gives equal corresponding angles and Â is common, so they are equiangular — and equiangular triangles are similar (AD : AB = AE : AC = 1 : 2).