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?)
Step-by-step solution
Idea: The Midpoint Theorem gives a line parallel to BS if both P and M are midpoints of the sides of ∆ABS. P is already the midpoint of AB, so make M the midpoint of AS.
- Placing S: extend AM beyond M to S with MS = AM, so M is the midpoint of AS.½ mark
- In ∆ABS, P is the midpoint of AB and M is the midpoint of AS, so PM ‖ BS (Midpoint Theorem). P, M, C lie on one line (the median CP), so MC ‖ BS.1 mark
- In ∆ACS, Q is the midpoint of AC and M is the midpoint of AS, so QM ‖ CS. Q, M, B lie on the median BQ, so MB ‖ CS. Yes, the same S gives the second pair of parallel sides.1 mark
- So BSCM has both pairs of opposite sides parallel: it is a parallelogram. Its diagonals BC and MS bisect each other at X, so X is the midpoint of BC. Hence AX (line AM) is the third median, and it passes through M: the medians are concurrent.1 mark
- The ratio: MX = ½MS = ½AM, so AM : MX = 2 : 1.½ mark
Answer to write in the exam
Produce AM to S with MS = AM (M midpoint of AS)
In ∆ABS: P, M midpoints of AB, AS ⇒ PM ‖ BS ⇒ MC ‖ BS (P, M, C collinear)
In ∆ACS: Q, M midpoints of AC, AS ⇒ QM ‖ CS ⇒ MB ‖ CS (Q, M, B collinear)
⇒ BSCM is a parallelogram ⇒ diagonals BC and MS bisect each other at X
⇒ X is the midpoint of BC ⇒ AX is the third median and passes through M
∴ Medians are concurrent; MX = ½MS = ½AM ⇒ AM : MX = 2 : 1.
Common mistakes that cost marks
- Choosing S on BC or making MS = MX. The construction works only with M the midpoint of AS.
- Proving only MC ‖ BS and calling BSCM a parallelogram. Both pairs of sides must be parallel (or one pair equal and parallel).
- Forgetting to conclude that AX is a median because X is the midpoint of BC.
How this can come in the exam
In the construction above, if AM = 8 cm, then MX equals
- 2 cm
- 4 cm
- 8 cm
- 16 cm
Show answer
(B) 4 cm
MS = AM = 8 cm and X is the midpoint of MS (diagonals of parallelogram BSCM), so MX = 4 cm.
Try one yourself
In the same figure, show that BS = 2MP.
Show answer
In ∆ABS, P and M are midpoints of AB and AS, so PM = ½BS by the Midpoint Theorem, i.e. BS = 2MP.
More questions like this
- 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.
- In a quadrilateral ABCD, suppose AB ‖ DC and AB ≠ CD. Suppose G and H are the midpoints of AC and BD respectively. Prove that GH ‖ AB. (Why did we assume AB ≠ CD?)
- Suppose the midpoints of sides AB, BC, CD and DA of a quadrilateral ABCD are P, Q, R and S respectively.
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