The segment joining the midpoints of two sides of a triangle is parallel to the third side and has half the length.
Step-by-step solution
To find: Prove that PQ ‖ BC and PQ = ½BC
Idea: Build a parallelogram that has BC as one side and PQ as half of the opposite side. The extra line through C parallel to BA does this: it copies AP across to the other side, where it equals BP.
- Construction. Through C draw line l parallel to BA. Extend PQ to meet l at R.½ mark
- In ∆APQ and ∆CRQ: AQ = CQ (Q is the midpoint of AC); ∠AQP = ∠CQR (vertically opposite angles); ∠APQ = ∠CRQ (alternate angles, AP ‖ RC). So ∆APQ ≅ ∆CRQ (AAS).1½ marks
- Hence PQ = QR, so PQ = ½PR; and CR = AP. Since AP = BP (P is the midpoint of AB), CR = BP.1 mark
- In quadrilateral BCRP, BP and CR are equal and parallel, so BCRP is a parallelogram. Therefore PR ‖ BC and PR = BC, which gives PQ ‖ BC and PQ = ½PR = ½BC.1 mark
Check: Coordinates: A(0, 6), B(−4, 0), C(8, 0). P = (−2, 3), Q = (4, 3). PQ is horizontal like BC, and PQ = 6 = ½ × 12 = ½BC ✓.
Answer to write in the exam
Construction: line l through C, l ‖ BA, meets PQ produced at R
In ∆APQ and ∆CRQ: AQ = CQ (Q midpoint of AC), ∠AQP = ∠CQR (vert. opp. angles), ∠APQ = ∠CRQ (alternate angles, AP ‖ RC)
∴ ∆APQ ≅ ∆CRQ (AAS) ⇒ PQ = QR and AP = CR (CPCT)
AP = BP ⇒ CR = BP; also CR ‖ BP ⇒ BCRP is a parallelogram
⇒ PR ‖ BC and PR = BC
∴ PQ ‖ BC and PQ = ½PR = ½BC.
Common mistakes that cost marks
- Joining CR without saying it is drawn parallel to BA. The parallel line is what makes the alternate angles equal.
- Writing the congruence as ∆APQ ≅ ∆RCQ. The correct matching is A↔C, P↔R, Q↔Q.
- Concluding BCRP is a parallelogram from CR = BP alone. The test needs a pair of sides that is equal AND parallel.
How this can come in the exam
In ∆XYZ, M and N are the midpoints of XY and XZ, and YZ = 13 cm. Then MN equals
- 13 cm
- 26 cm
- 6.5 cm
- 4.33 cm
Show answer
(C) 6.5 cm
By the Midpoint Theorem, MN = ½YZ = 6.5 cm.
A gardener makes a triangular flower bed ABC with BC = 18 m. She ties a rope from the midpoint P of side AB to the midpoint Q of side AC to separate roses (near A) from marigolds.
(i) How long is the rope PQ? (ii) Is the rope parallel to BC? Give the theorem. (iii) The rose part APQ has perimeter 21 m; AB = 16 m. Find AC. (iv) What fraction of the bed’s area is roses?
Show answer
(i) PQ = ½ × 18 = 9 m (1 mark). (ii) Yes, by the Midpoint Theorem PQ ‖ BC (1 mark). (iii) AP + AQ + PQ = 21 ⇒ 8 + AQ + 9 = 21 ⇒ AQ = 4 ⇒ AC = 8 m (1 mark). (iv) ∆APQ is one of four congruent triangles made by the three midpoints, so roses take ¼ of the area (1 mark).Try one yourself
In ∆ABC, P and Q are the midpoints of AB and AC. If ∠ABC = 62°, find ∠APQ.
Show answer
PQ ‖ BC (Midpoint Theorem), so ∠APQ = ∠ABC = 62° (corresponding angles).
More questions like this
- Read the statement of the Midpoint Theorem again. The following questions seem natural to ask. What can we say about (1) a segment PQ (with P on side AB and Q on side AC) that is parallel to the third side and has half the length, and (2) the line that passes through the midpoint of one side and is parallel to another side? For now, experiment and see if you can make a guess about question (2).
- The line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side (and the resulting segment has half the length of the parallel side).
- BCRP is a parallelogram, so CR = BP. BP = PA and ∠AQP = ∠CQR (vertically opposite angles). Combining with CR ‖ BA, ∆APQ ≅ ∆CRQ. (Why?)
- We will later generalise the converse of the Midpoint Theorem by dropping the condition that P is the midpoint of AB and assuming only that PQ ‖ BC. Can you guess the conclusion in this case? (Hint: the name of the more general theorem includes the word proportionality.)
- In ∆ABC, suppose P, Q and R are the midpoints of side AB, AC and BC respectively. Instead of joining the midpoints with each other as we did earlier, draw segments AR, BQ and CP. These are called the medians of ∆ABC. What do you see? Repeat this for several triangles and see that the three medians always seem to be concurrent, meaning they pass through a common point! There is more! Assume for the moment that the medians are always concurrent. Let M be the common point where they meet. Compare the lengths of the two parts CM and MP of the median CP, and do the same for the other two medians. Do you see a pattern?
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