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?).
Answer: PQ ‖ XY, so with transversal PY (the median CP) ∠MPQ = ∠MYX, and with transversal QX (the median BQ) ∠MQP = ∠MXY (alternate angles). The side between these angles is PQ = YX (both ½BC). So ∆MPQ ≅ ∆MYX by ASA.
Step-by-step solution
Idea: ASA needs two angles and the side between them. Parallel lines PQ and XY give the angles; the Midpoint Theorem (used twice) gives the equal side.
- P, M, Y lie on the median CP, and Q, M, X lie on the median BQ. These two lines are transversals of the parallel lines PQ and XY.½ mark
- Alternate angles: ∠MPQ = ∠MYX (transversal PY) and ∠MQP = ∠MXY (transversal QX).1 mark
- The included sides: PQ = ½BC (Midpoint Theorem in ∆ABC) and YX = ½BC (Midpoint Theorem in ∆MBC), so PQ = YX. Hence ∆MPQ ≅ ∆MYX (ASA).½ mark
∆MPQ ≅ ∆MYX by ASA: ∠MPQ = ∠MYX, PQ = YX, ∠MQP = ∠MXY (alternate angles, since PQ ‖ XY).
Answer to write in the exam
PQ ‖ BC and XY ‖ BC (Midpoint Theorem in ∆ABC, ∆MBC) ⇒ PQ ‖ XY
∠MPQ = ∠MYX and ∠MQP = ∠MXY (alternate angles)
PQ = ½BC = YX
∴ ∆MPQ ≅ ∆MYX (ASA)
Common mistakes that cost marks
- Matching P with X and Q with Y. The correct correspondence is P↔Y, Q↔X (the triangles are turned round through M).
- Using vertically opposite angles at M with two non-included angles and calling it ASA; for ASA the equal side must lie between the two equal angles.
- Forgetting to say why PQ = XY: both equal ½BC.
How this can come in the exam
MCQ (1 mark)
From ∆MPQ ≅ ∆MYX (with P↔Y, Q↔X), which equality follows?
- MP = MX
- MQ = MX
- PQ = MY
- MP = XY
Show answer
(B) MQ = MX
Corresponding sides: MP = MY, MQ = MX, PQ = YX. So MQ = MX.
Try one yourself
Using MQ = MX and the fact that X is the midpoint of BM, show that BM : MQ = 2 : 1.
Show answer
BX = XM = MQ, so BM = BX + XM = 2MQ, i.e. BM : MQ = 2 : 1.
More questions like this
- 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.
- Give another proof of the Centroid Theorem using a clever construction that we will describe only partly and ask you to complete. Once again let M be the intersection point of the two medians CP and BQ. We want to show that the intersection point of line AM with side BC is the midpoint of BC. For this, extend AM up to a carefully chosen point S such that BSCM becomes a parallelogram. (Hint: To arrange MC ‖ BS, first note that PM and MC are along the same line. Now where should M be on segment AS? Does your placement of S also prove MB ‖ CS?)
- The midpoints of the four sides of a quadrilateral are the vertices of a parallelogram.
- (i) If P, Q, R are the midpoints of sides AB, AC, BC respectively of ∆ABC, show that ∆PQR is congruent to ∆QPA and to two other triangles which you should identify.
(ii) Suppose someone erases ∆ABC, leaving only ∆PQR on the paper. Can you reconstruct ∆ABC from ∆PQR? - In ∆ABC, let M and N be midpoints of AB and AC respectively. Let D be any point on BC. Show that MN bisects AD.
All Quadrilaterals and parallelograms questions · All maths questions