Multiple converses to a theorem. Let us see how multiple statements can be considered converses to the Midpoint Theorem and how the converse of the Midpoint Theorem (the line through the midpoint of one side parallel to another side bisects the third side) is one of them. To formulate a converse we should express the original statement in “If … then …” form. For a complex statement, there may be multiple ways to do that. The Midpoint Theorem starts with ∆ABC and points P and Q on sides AB and AC respectively. The theorem has two assumptions and two conclusions, which we have named for further discussion.
Assumptions: (P MID) P is the midpoint of AB, and (Q MID) Q is the midpoint of AC.
Conclusions: (PRLL) PQ ‖ BC, and (HALF) PQ = BC2.
Let us use these four named conditions to discuss various possible statements.
- (i) A natural converse of the Midpoint Theorem would be: “If (PRLL) and (HALF) are true, then (P MID) and (Q MID) are true.” Write out this statement fully. Experiment and see that it seems to be true! Prove the statement. (Hint: Try to run a proof of the Midpoint Theorem backwards.)
- (ii) To see why the converse of the Midpoint Theorem really is a converse of the Midpoint Theorem, we first write the Midpoint Theorem as follows: “Suppose P is the midpoint of side AB of ∆ABC and Q is a point on side AC. If Q is the midpoint of AC then PQ ‖ BC.” (Also PQ = BC2, but let us set that aside for now.) Write the converse of the second sentence in quotes by keeping the first sentence the same. Verify that the converse of the Midpoint Theorem is what you get! (And PQ = BC2 also follows.) Now write it in terms of the four named conditions.
- (iii) The discussion so far suggests that it is reasonable to take two of the four listed statements as assumptions and ask if the other two are true. Verify that only one such combination remains to be examined. “If (P MID) and (HALF) are true, then can we conclude (Q MID) and/or (PRLL)?” Write this out in words. This is a precise version of question (1) stated before the converse of the Midpoint Theorem. Can you answer it?
Answer: (i) “In ∆ABC with P on AB and Q on AC, if PQ ‖ BC and PQ = ½BC, then P and Q are the midpoints of AB and AC.” True. (ii) “Suppose P is the midpoint of side AB of ∆ABC and Q is a point on side AC. If PQ ‖ BC, then Q is the midpoint of AC”: this is the converse of the Midpoint Theorem, i.e. (P MID) and (PRLL) ⇒ (Q MID) (and (HALF)). (iii) The only remaining pair is (P MID) and (HALF); no, neither (Q MID) nor (PRLL) need follow: e.g. A(0, 0), B(3, 3), C(4, 0), P(1.5, 1.5) and Q(1, 0) has PQ = ½BC but Q is not the midpoint and PQ ∦ BC.
Step-by-step solution
Idea: Take any two of the four conditions as the “if” part and ask whether the other two follow. Swapping the roles of P and Q gives nothing new, which leaves just one new case.
(i) A natural converse of the Midpoint Theorem would be: “If (PRLL) and (HALF) are true, then (P MID) and (Q MID) are true.” Write out this statement fully. Experiment and see that it seems to be true! Prove the statement. (Hint: Try to run a proof of the Midpoint Theorem backwards.)
- Statement: In ∆ABC, let P be a point on AB and Q a point on AC. If PQ ‖ BC and PQ = ½BC, then P is the midpoint of AB and Q is the midpoint of AC.½ mark
- Proof: extend PQ beyond Q to R with QR = PQ. Then PR = 2PQ = BC and PR ‖ BC, so BCRP is a parallelogram: CR = BP and CR ‖ BA.½ mark
- In ∆APQ and ∆CRQ: PQ = RQ, ∠AQP = ∠CQR (vertically opposite), ∠APQ = ∠CRQ (alternate angles, AP ‖ CR). So ∆APQ ≅ ∆CRQ (AAS): AQ = CQ and AP = CR = BP. Hence P and Q are midpoints.1 mark
True: if PQ ‖ BC and PQ = ½BC, then P and Q are the midpoints of AB and AC.
(ii) To see why the converse of the Midpoint Theorem really is a converse of the Midpoint Theorem, we first write the Midpoint Theorem as follows: “Suppose P is the midpoint of side AB of ∆ABC and Q is a point on side AC. If Q is the midpoint of AC then PQ ‖ BC.” (Also PQ = BC2, but let us set that aside for now.) Write the converse of the second sentence in quotes by keeping the first sentence the same. Verify that the converse of the Midpoint Theorem is what you get! (And PQ = BC2 also follows.) Now write it in terms of the four named conditions.
- Converse of the second sentence: “If PQ ‖ BC, then Q is the midpoint of AC.” With the first sentence kept: “Suppose P is the midpoint of side AB of ∆ABC and Q is a point on side AC. If PQ ‖ BC, then Q is the midpoint of AC.” This says the line through the midpoint of AB parallel to BC bisects AC: the converse of the Midpoint Theorem (and then PQ = ½BC by the Midpoint Theorem).1 mark
- In named conditions: if (P MID) and (PRLL), then (Q MID) and (HALF).½ mark
(P MID) and (PRLL) ⇒ (Q MID) and (HALF).
(iii) The discussion so far suggests that it is reasonable to take two of the four listed statements as assumptions and ask if the other two are true. Verify that only one such combination remains to be examined. “If (P MID) and (HALF) are true, then can we conclude (Q MID) and/or (PRLL)?” Write this out in words. This is a precise version of question (1) stated before the converse of the Midpoint Theorem. Can you answer it?
- Pairs of assumptions: {P MID, Q MID} (the theorem), {PRLL, HALF} (part (i)), {P MID, PRLL} (part (ii)); {Q MID, PRLL} is (ii) with P and Q swapped; {Q MID, HALF} is {P MID, HALF} with P and Q swapped. So only {P MID, HALF} is new. In words: “P is the midpoint of AB, Q is on AC, and PQ = ½BC. Must Q be the midpoint of AC? Must PQ be parallel to BC?”½ mark
- Answer: not necessarily. Points of AC at distance ½BC from P lie on a circle centred at P; it passes through the midpoint M of AC but can cross AC at a second point. Example: A(0, 0), B(3, 3), C(4, 0); P = (1.5, 1.5), ½BC = ½√10 = √2.5. Q = (1, 0) is on AC and PQ = √(0.25 + 2.25) = √2.5, yet Q is not the midpoint (2, 0) and PQ is not parallel to BC. So neither (Q MID) nor (PRLL) follows.1 mark
No: with P the midpoint and PQ = ½BC, Q may be a second point of AC (not the midpoint), and then PQ is not parallel to BC.
(i) (PRLL) and (HALF) ⇒ (P MID) and (Q MID): true. (ii) (P MID) and (PRLL) ⇒ (Q MID) and (HALF): the converse of the Midpoint Theorem. (iii) (P MID) and (HALF) do not force (Q MID) or (PRLL): a second point of AC at distance ½BC from P is a counterexample.
Check: (iii) Slopes: PQ from (1.5, 1.5) to (1, 0) has slope 3; BC from (3, 3) to (4, 0) has slope −3, so not parallel ✓. PQ² = 0.25 + 2.25 = 2.5 and (½BC)² = 10/4 = 2.5 ✓.
Answer to write in the exam
(i)
Produce PQ to R, QR = PQ ⇒ PR = BC, PR ‖ BC ⇒ BCRP parallelogram ⇒ CR = BP, CR ‖ BA
∆APQ ≅ ∆CRQ (AAS: PQ = RQ, ∠AQP = ∠CQR, ∠APQ = ∠CRQ)
⇒ AQ = CQ and AP = CR = BP
∴ (PRLL) and (HALF) ⇒ (P MID) and (Q MID)
(ii)
Converse of second sentence: If PQ ‖ BC, then Q is the midpoint of AC
With P the midpoint of AB: this is the converse of the Midpoint Theorem
∴ (P MID) and (PRLL) ⇒ (Q MID) and (HALF)
(iii)
Six pairs; by the P–Q symmetry only (P MID) and (HALF) is new
A(0, 0), B(3, 3), C(4, 0): P(1.5, 1.5), BC = √10
Q(1, 0) on AC: PQ = √2.5 = ½BC, but midpoint of AC is (2, 0) and PQ ∦ BC
∴ (P MID) and (HALF) do not imply (Q MID) or (PRLL)
Common mistakes that cost marks
- Counting six cases and examining all of them; four are covered already or are mirror images (P ↔ Q).
- In (iii), answering “yes” from a drawing where the circle happens to meet AC only at the midpoint.
- In (i), using the Midpoint Theorem itself (which assumes the midpoints) instead of proving the new statement.
How this can come in the exam
Assertion–Reason (1 mark)
Assertion (A): In ∆ABC, if P is the midpoint of AB and Q is a point of AC with PQ = ½BC, then Q must be the midpoint of AC.
Reason (R): In ∆ABC, the segment joining the midpoints of AB and AC is half of BC.
- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Show answer
(D) A is false but R is true.
R is the Midpoint Theorem (true). A is false: another point of AC can be at the same distance ½BC from P.
Try one yourself
In ∆ABC, P is a point on AB and Q a point on AC with PQ ‖ BC and PQ = 4 cm. If BC = 8 cm and AB = 10 cm, find AP.
Show answer
PQ ‖ BC and PQ = ½BC, so P and Q are midpoints: AP = 5 cm.
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