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).
Step-by-step solution
To find: Prove that AQ = QC and PQ = ½BC
Idea: Run the proof of the Midpoint Theorem backwards. This time PQ ‖ BC is given, so BCRP is a parallelogram at once (two pairs of parallel sides), and the congruent triangles give the midpoint of AC.
- Construction. Through C draw the line parallel to BA; let it meet line PQ at R.½ mark
- PR ‖ BC (given) and CR ‖ BP (construction), so BCRP is a parallelogram. Hence CR = BP and PR = BC. Also BP = PA (P is the midpoint), so CR = PA.1 mark
- In ∆APQ and ∆CRQ: AP = CR; ∠PAQ = ∠RCQ (alternate angles, AB ‖ CR, transversal AC); ∠AQP = ∠CQR (vertically opposite). So ∆APQ ≅ ∆CRQ (AAS).1½ marks
- Hence AQ = QC, so Q is the midpoint of AC: the line bisects the third side. Also PQ = QR, so PQ = ½PR = ½BC.1 mark
Check: Second proof: if M is the midpoint of AC, then PM ‖ BC by the Midpoint Theorem. Through P there is only one line parallel to BC, so line PQ is line PM and Q = M ✓.
Answer to write in the exam
Construction: line through C ‖ BA meets PQ produced at R
PR ‖ BC (given), CR ‖ BP (construction) ⇒ BCRP is a parallelogram ⇒ CR = BP = PA, PR = BC
In ∆APQ and ∆CRQ: AP = CR, ∠PAQ = ∠RCQ (alternate angles), ∠AQP = ∠CQR (vert. opp. angles)
∴ ∆APQ ≅ ∆CRQ (AAS) ⇒ AQ = QC and PQ = QR (CPCT)
∴ Q is the midpoint of AC and PQ = ½PR = ½BC.
Common mistakes that cost marks
- Using the Midpoint Theorem’s conclusion (Q is the midpoint) as a reason. Here it must be proved.
- Forgetting to show CR = PA before claiming the congruence.
- Calling ∠PAQ and ∠RCQ corresponding angles; they are alternate angles on transversal AC.
How this can come in the exam
In ∆ABC, D is the midpoint of AB and DE ‖ BC with E on AC. If AE = 4.2 cm and BC = 9 cm, find AC and DE.
Show answer
By the converse of the Midpoint Theorem, E is the midpoint of AC, so AC = 2 × 4.2 = 8.4 cm (1 mark), and DE = ½BC = 4.5 cm (1 mark).Try one yourself
ABCD is a trapezium with AB ‖ DC. E is the midpoint of AD, and the line through E parallel to AB meets the diagonal BD at G. Where is G?
Show answer
In ∆DAB, E is the midpoint of DA and EG ‖ AB, so G is the midpoint of BD (converse of the Midpoint Theorem).
More questions like this
- 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?
- Let Y and X be the midpoints of CM and BM respectively. Now we can use the Midpoint Theorem in ∆ABC and in ∆MBC! We get that PQ and XY are parallel to BC and hence to each other. Moreover PQ = XY = BC2. So ∆MPQ ≅ ∆MYX by ASA (How?).
- The three medians of ∆ABC pass through a common point, which divides each of the medians in the ratio 2 : 1, with the longer part connecting to the vertex. This point is called the centroid of ∆ABC.
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