We will later generalise the converse of the Midpoint Theorem by dropping the condition that P is the midpoint of AB and assuming only that PQ ‖ BC. Can you guess the conclusion in this case? (Hint: the name of the more general theorem includes the word proportionality.)

Step-by-step solution

Idea: When P is the midpoint, AP : PB = 1 : 1 and the converse of the Midpoint Theorem says AQ : QC = 1 : 1 too. The natural guess is that whatever ratio P makes on AB, Q makes the same ratio on AC.

1. Midpoint case: AP : PB = 1 : 1 gives AQ : QC = 1 : 1. Experiment with P one-third of the way along AB: the parallel line meets AC one-third of the way along too.1 mark
2. Guess: a line parallel to one side of a triangle divides the other two sides in the same ratio: APPB = AQQC. This is the Basic Proportionality Theorem.1 mark

Answer to write in the exam

Common mistakes that cost marks

• Guessing PQ = ½BC. That needs P to be the midpoint; in general PQ : BC = AP : AB.
• Writing AP/PB = AQ/AC, mixing a part with a whole side.

How this can come in the exam

Try one yourself

In ∆XYZ, a line parallel to YZ meets XY at M and XZ at N. XM = 4, MY = 6, XN = 5. Find NZ.

More questions like this

• 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?
• 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.

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