Suppose P is a point on side AB of ∆ABC and the line through P parallel to BC meets AC in point Q. For parts (i) to (iii), assume AP = 1, AQ = √5, and see the figure.
- (i) If PB = 2 find QC. (Hint: See the dotted line)
- (ii) If PB = 13 find QC. (Hint: See the dotted lines)
- (iii) If PB = 23 find QC. (Hint: Divide both PB and AP suitably.)
- (iv) Show that if APPB is a rational number, then APPB = AQQC. Later, you will prove this equality without assuming APPB is rational. That will require a new idea.
Step-by-step solution
Idea: Key fact: parallel lines that cut equal pieces from one side of a triangle cut equal pieces from the other side. It follows from the converse of the Midpoint Theorem (or from congruent triangles), and it turns length ratios on AB into the same ratios on AC.
(i) If PB = 2 find QC. (Hint: See the dotted line)
- Let X be the midpoint of PB, so AP = PX = XB = 1, and let the line through X parallel to BC meet AC at Y. In ∆AXY, P is the midpoint of AX and PQ ‖ XY, so Q is the midpoint of AY: QY = AQ = √5. In trapezium PBCQ (PQ ‖ BC), X is the midpoint of PB and XY ‖ BC, so Y is the midpoint of QC (join PC; use the converse of the Midpoint Theorem in ∆PBC and then in ∆CPQ): YC = QY = √5. So QC = 2√5.1 mark
(ii) If PB = 13 find QC. (Hint: See the dotted lines)
- Divide AP into three parts of 13 each. Then AB is cut into 4 equal parts of 13 (three in AP and PB itself). Lines through the cut points parallel to BC cut AC into 4 equal parts; AQ = √5 is three of them, so each is √53 and QC = √53.1 mark
(iii) If PB = 23 find QC. (Hint: Divide both PB and AP suitably.)
- Use parts of length 13: AP = 3 parts, PB = 2 parts. The parallels to BC through the 4 cut points divide AC into 5 equal parts, with AQ = 3 parts = √5. One part = √53, so QC = 2 parts = 2√53.1 mark
(iv) Show that if APPB is a rational number, then APPB = AQQC. Later, you will prove this equality without assuming APPB is rational. That will require a new idea.
- Equal steps give equal steps. Let A = P0, P1, …, Pk = B be equally spaced on AB, and let the lines through them parallel to BC meet AC at Q0 = A, Q1, …, Qk = C. Through each Qi draw a line parallel to AB, meeting the next parallel at Ri. Each PiPi+1RiQi is a parallelogram, so QiRi = PiPi+1, all equal. The triangles QiRiQi+1 are all congruent (ASA: equal side, angle = ∠A and angle = ∠B by corresponding angles), so all the pieces QiQi+1 are equal.1½ marks
- If APPB = mn (whole numbers), cut AB into m + n equal parts so that P is the m-th cut point. Then AC is cut into m + n equal parts with Q at the m-th point, so AQQC = mn = APPB.½ mark
Check: Each answer satisfies AP/PB = AQ/QC: (i) 1/2 = √5/(2√5) ✓; (ii) 1/(1/3) = 3 = √5/(√5/3) ✓; (iii) 1/(2/3) = 3/2 = √5/(2√5/3) ✓.
Answer to write in the exam
(i)
X midpoint of PB (PX = XB = 1); XY ‖ BC, Y on AC
∆AXY: P midpoint of AX, PQ ‖ XY ⇒ QY = AQ = √5
Trapezium PBCQ: X midpoint of PB, XY ‖ BC ⇒ Y midpoint of QC ⇒ YC = √5
∴ QC = 2√5
(ii)
AP = 1 = 3 × ⅓, PB = ⅓ ⇒ AB in 4 equal parts of ⅓
Parallels to BC ⇒ AC in 4 equal parts; AQ = 3 parts = √5 ⇒ 1 part = √5/3
∴ QC = √5/3
(iii)
AP = 3 × ⅓, PB = 2 × ⅓ ⇒ AB in 5 equal parts
AC in 5 equal parts; AQ = 3 parts = √5 ⇒ 1 part = √5/3
∴ QC = 2√5/3
(iv)
Let AP/PB = m/n; divide AB into m + n equal parts, P at the m-th point
Lines ‖ BC through the points divide AC into m + n equal parts (parallelograms + congruent triangles, ASA)
Q is at the m-th point of AC ⇒ AQ/QC = m/n
∴ AP/PB = AQ/QC
Common mistakes that cost marks
- Assuming QC = PB × √5 without justification; the reason is the equal-parts argument.
- In (iii), dividing PB into 2 parts and AP into 1 part: the parts must all be the same length (⅓ each).
- Writing AP/AB = AQ/QC, mixing a whole side with a part.
How this can come in the exam
In ∆ABC, PQ ‖ BC with P on AB and Q on AC. AP = 2, PB = 3 and AQ = 4. Then QC is
- 5
- 6
- 8/3
- 1.5
Show answer
(B) 6
AP/PB = AQ/QC ⇒ 2/3 = 4/QC ⇒ QC = 6.
Try one yourself
In ∆ABC, PQ ‖ BC with P on AB, Q on AC, AP = 3, PB = 1 and QC = 2.5. Find AQ.
Show answer
AQ/QC = AP/PB = 3 ⇒ AQ = 7.5.
More questions like this
- (i) Suppose ABCD is a parallelogram and M, N are midpoints of AB and CD respectively. Show that segments DM and BN trisect segment AC.
(ii) Use part (i) to find a procedure to trisect any given segment PQ. Find two other ways to trisect PQ, one using the fact that a line parallel to one side of a triangle divides the other two sides in the same ratio and a third using the Centroid Theorem. - Is the Midpoint Theorem for Quadrilaterals true when the quadrilateral is non-convex? How about when it is self-intersecting? Experiment and check. The pictures may look strange, but the theorem appears to be true.
- Let P, Q, R, S be four points on sides AB, BC, CD and DA respectively of a quadrilateral ABCD. Suppose PQRS is a parallelogram. Must P, Q, R and S be midpoints of the respective sides? This can be considered a possible converse question to the Midpoint Theorem for Quadrilaterals.
- (i) Complete the following proof of the Midpoint Theorem. Extend segment PQ beyond Q until point S. By how much should we extend PQ? It would be good to be able to prove that ∆APQ ≅ ∆CSQ. (Why?) Use this as a guide to specify the location of S and then complete this proof.
(ii) Give a similar proof of the converse of the Midpoint Theorem. Start by extending PQ up to a suitable point S. - Review all the properties of a rhombus/rectangle/square that you proved earlier. Formulate a converse of each. Decide if the converse is true. There are many possibilities here!
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