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?

Step-by-step solution

Idea: Measure carefully on a few very different triangles. The same two facts appear every time, which suggests a theorem (the Centroid Theorem).

1. Draw the three medians in several triangles (acute, right, obtuse, long and thin). In each case the three medians meet at a single point M: they are concurrent.1 mark
2. Example measurement: in a triangle with median CP = 9 cm, CM = 6 cm and MP = 3 cm. On the other medians the same happens: the vertex part is twice the midpoint part.1 mark
3. Pattern: CM : MP = 2 : 1, and likewise AM : MR = 2 : 1 and BM : MQ = 2 : 1. So M is two-thirds of the way along each median from the vertex.1 mark

Check: Coordinates: A(0, 6), B(−6, 0), C(6, 0). Medians meet at M(0, 2), the average of the vertices. On median CP with P(−3, 3): CM = √(36 + 4) = √40 and MP = √(9 + 1) = √10, and √40 = 2√10 ✓.

Answer to write in the exam

Common mistakes that cost marks

• Joining a vertex to the midpoint of an adjacent side. A median goes to the midpoint of the opposite side.
• Writing the ratio the wrong way round (1 : 2). The longer part is next to the vertex.
• Drawing medians by eye: mark midpoints by measuring or folding, or the lines will not meet at one point.

How this can come in the exam

Try one yourself

The centroid G of ∆PQR lies on median PS with GS = 2.5 cm. Find PG and PS.

More questions like this

• 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?).
• 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?

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