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).
- (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
- (2) the line that passes through the midpoint of one side and is parallel to another side
Step-by-step solution
Idea: Both questions turn the Midpoint Theorem around: what was the conclusion becomes the assumption. Experiment with a ruler, then prove each one with the same parallelogram trick used for the Midpoint Theorem.
(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
- Extend PQ beyond Q to R with QR = PQ. Then PR = 2PQ = BC and PR ‖ BC, so BCRP is a parallelogram (one pair of opposite sides equal and parallel). Hence CR = BP and CR ‖ BA.1 mark
- In ∆APQ and ∆CRQ: PQ = RQ, ∠APQ = ∠CRQ (alternate angles, AP ‖ CR), ∠AQP = ∠CQR (vertically opposite). So ∆APQ ≅ ∆CRQ (AAS): AQ = CQ and AP = CR = BP. P and Q are the midpoints.1 mark
(2) the line that passes through the midpoint of one side and is parallel to another side
- Guess from experiment: draw ∆ABC, mark the midpoint P of AB, and draw the line through P parallel to BC. Measure: it meets AC at Q with AQ = QC every time. So the line bisects the third side.1 mark
- Reason: let M be the midpoint of AC. By the Midpoint Theorem PM ‖ BC. Only one line through P is parallel to BC, so line PQ is line PM, and it meets AC at M. Hence Q = M, the midpoint, and PQ = ½BC.1 mark
Answer to write in the exam
(1)
Produce PQ to R with QR = PQ ⇒ PR = BC, PR ‖ BC ⇒ BCRP parallelogram ⇒ CR = BP, CR ‖ BA
∆APQ ≅ ∆CRQ (AAS: PQ = RQ, ∠APQ = ∠CRQ, ∠AQP = ∠CQR) ⇒ AQ = QC, AP = CR = BP
∴ P and Q are midpoints of AB and AC.
(2)
P midpoint of AB, line through P ‖ BC meets AC at Q; M = midpoint of AC
PM ‖ BC (Midpoint Theorem); only one line through P ‖ BC ⇒ line PQ = line PM
∴ Q = M: the line bisects AC, and PQ = ½BC.
Common mistakes that cost marks
- In (2), thinking the line bisects the side it is parallel to. It bisects the third side.
- In (1), using only “parallel” or only “half the length”. Both conditions together force the midpoints.
- Guessing from one drawing only. Try several triangles before trusting the pattern.
How this can come in the exam
In ∆PQR, S is the midpoint of PQ and the line through S parallel to QR meets PR at T. If PR = 11 cm, then PT is
- 5.5 cm
- 11 cm
- 22 cm
- 3.67 cm
Show answer
(A) 5.5 cm
The line through the midpoint of PQ parallel to QR bisects PR, so PT = 11 ÷ 2 = 5.5 cm.
Try one yourself
In ∆ABC, D is the midpoint of BC. A line through D parallel to BA meets AC at E. If AB = 9 cm and AC = 14 cm, find DE and AE.
Show answer
The line bisects AC, so E is the midpoint of AC and AE = 7 cm; DE = ½AB = 4.5 cm.
More questions like this
- 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?
- 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?).
All Quadrilaterals and parallelograms questions · All maths questions