Midpoint theorem · 4 marks

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?)

Answer: Let K be the midpoint of AD. In ∆ADC, KG ‖ DC; in ∆DAB, KH ‖ AB ‖ DC. Only one line through K is parallel to DC, so K, H, G are collinear and GH ‖ AB. If AB = CD, ABCD would be a parallelogram, whose diagonals bisect each other, so G = H and there is no segment GH.

Step-by-step solution

Given: ABCD with AB ‖ DC, AB ≠ CD; G, H midpoints of AC, BD
To find: Prove that GH ‖ AB, and explain why AB ≠ CD is assumed

Idea: Bring in a third midpoint, K on AD. The Midpoint Theorem in two triangles that share side AD gives two segments from K, both parallel to the parallel sides. They must lie on one line.

Let K be the midpoint of AD. In ∆ADC, K and G are midpoints of AD and AC, so KG ‖ DC (Midpoint Theorem).1 mark
In ∆DAB, K and H are midpoints of DA and DB, so KH ‖ AB. Since AB ‖ DC, KH ‖ DC.1 mark
Through K there is only one line parallel to DC, so KG and KH lie on the same line. So K, H, G are collinear and GH lies along this line: GH ‖ DC ‖ AB.1 mark
Why AB ≠ CD? If AB = CD (with AB ‖ DC), ABCD is a parallelogram (equal and parallel opposite sides). Its diagonals bisect each other, so G and H are the same point and “GH” is not a segment at all.1 mark

GH ‖ AB (G and H lie on the line through the midpoint of AD parallel to the parallel sides). AB ≠ CD is assumed because otherwise G = H.

Check: A(0, 4), B(4, 4), C(10, 0), D(0, 0): G = (5, 2), H = (2, 2). GH is horizontal like AB ✓, and GH = 3 = (10 − 4)/2 = (DC − AB)/2.

Answer to write in the exam

Let K be the midpoint of AD

In ∆ADC: K, G midpoints of AD, AC ⇒ KG ‖ DC (Midpoint Theorem)

In ∆DAB: K, H midpoints of DA, DB ⇒ KH ‖ AB ‖ DC

Only one line through K ‖ DC ⇒ K, H, G collinear

∴ GH ‖ AB.

If AB = CD, ABCD is a parallelogram ⇒ diagonals bisect each other ⇒ G = H; so AB ≠ CD is needed.

Common mistakes that cost marks

Applying the Midpoint Theorem in ∆ABC: G is the midpoint of AC, but H is not on a side of ∆ABC.
Showing KG ‖ DC and KH ‖ AB, then stopping. The step “only one line through K is parallel to DC, so K, H, G are collinear” is essential.
Answering the “why” part with “so that it is a trapezium”. The real reason is that G and H would coincide.

How this can come in the exam

MCQ (1 mark)

ABCD is a trapezium with AB ‖ DC, AB = 6 cm and DC = 14 cm. G and H are the midpoints of the diagonals AC and BD. Then GH equals

4 cm
8 cm
10 cm
20 cm

Show answer

(A) 4 cm
With K the midpoint of AD: KG = ½DC = 7 cm and KH = ½AB = 3 cm along the same line, so GH = 7 − 3 = 4 cm.

Try one yourself

In trapezium PQRS with PQ ‖ SR, PQ = 5 cm and SR = 11 cm. Find the distance between the midpoints of the diagonals PR and QS.