A more general midpoint theorem and its converse. In a quadrilateral ABCD, suppose AB ‖ DC. Recall that such ABCD is called a trapezium. Let E be the midpoint of AD. A line drawn through E intersects side BC at F.
- (i) If EF ‖ AB, then show that F is the midpoint of BC. Conclude that EF = (AB + CD)2.
- (ii) If F is the midpoint of BC, then show that EF ‖ AB. There are at least two ways to solve (ii). Do it directly by using the midpoint M of BD and showing that EM and FM are the same lines. Or use (i) along with the same idea used in the second proof of the converse of the Midpoint Theorem. Which way do you think is simpler?
Step-by-step solution
Idea: A diagonal splits the trapezium into two triangles, each having one of the parallel sides as a side. Apply the Midpoint Theorem (or its converse) in each triangle.
(i) If EF ‖ AB, then show that F is the midpoint of BC. Conclude that EF = (AB + CD)2.
- Draw diagonal BD; let EF meet it at G. In ∆ABD, E is the midpoint of AD and EG ‖ AB, so G is the midpoint of BD and EG = ½AB (converse of the Midpoint Theorem).1 mark
- EF ‖ AB ‖ DC, so GF ‖ DC. In ∆BDC, G is the midpoint of BD and GF ‖ DC, so GF bisects BC: F is the midpoint of BC, and GF = ½DC.1 mark
- EF = EG + GF = ½AB + ½DC = AB + CD2.1 mark
(ii) If F is the midpoint of BC, then show that EF ‖ AB. There are at least two ways to solve (ii). Do it directly by using the midpoint M of BD and showing that EM and FM are the same lines. Or use (i) along with the same idea used in the second proof of the converse of the Midpoint Theorem. Which way do you think is simpler?
- Direct way: let M be the midpoint of BD. In ∆DAB, E and M are midpoints of DA and DB, so EM ‖ AB. In ∆BCD, F and M are midpoints of BC and BD, so FM ‖ DC ‖ AB.1 mark
- Through M there is only one line parallel to AB, so EM and FM are the same line: E, M, F are collinear and EF ‖ AB. (Second way: the line through E parallel to AB meets BC at its midpoint by (i); that midpoint is F, so this line is EF.) The direct way is simpler: it uses only the Midpoint Theorem twice.1 mark
Check: A(0, 4), B(4, 4), C(10, 0), D(0, 0): E(0, 2), F(7, 2); EF = 7 and (AB + CD)/2 = (4 + 10)/2 = 7 ✓; EF is horizontal like AB ✓.
Answer to write in the exam
(i)
Join BD, meeting EF at G
∆ABD: E midpoint of AD, EG ‖ AB ⇒ G midpoint of BD, EG = ½AB (converse of Midpoint Theorem)
∆BDC: G midpoint of BD, GF ‖ DC ⇒ F midpoint of BC, GF = ½DC
∴ EF = EG + GF = (AB + CD)/2
(ii)
M = midpoint of BD
∆DAB: EM ‖ AB; ∆BCD: FM ‖ DC ‖ AB (Midpoint Theorem)
Only one line through M ‖ AB ⇒ E, M, F collinear
∴ EF ‖ AB
Common mistakes that cost marks
- Writing EF = ½(AB − CD) or ½AB. The two pieces EG and GF add up.
- Using the Midpoint Theorem in ∆ABC with E: E is not on a side of ∆ABC.
- In (ii), assuming E, M, F lie on one line before proving it.
How this can come in the exam
In trapezium PQRS with PQ ‖ SR, PQ = 9 cm and SR = 15 cm. The segment joining the midpoints of PS and QR has length
- 3 cm
- 6 cm
- 12 cm
- 24 cm
Show answer
(C) 12 cm
It equals ½(9 + 15) = 12 cm.
The segment joining the midpoints of the non-parallel sides of a trapezium is 11 cm, and one parallel side is 8 cm. Find the other parallel side.
Show answer
½(8 + x) = 11 (1 mark) ⇒ x = 14 cm (1 mark).Try one yourself
A ladder-shaped trapezium has parallel rungs of 40 cm (top) and 64 cm (bottom). A middle rung joins the midpoints of the slanting sides. How long is it?
Show answer
½(40 + 64) = 52 cm.
More questions like this
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- Consider 4 points A, B, C, D in the plane with no three collinear. Answer the following questions. Some answers may require you to consider different cases, depending on how the points are positioned in the plane.
- Suppose P is a point on side AB of ∆ABC and the line through P parallel to BC meets AC in point Q. For parts (i) to (iii), assume AP = 1, AQ = √5, and see the figure.
- (i) Suppose ABCD is a parallelogram and M, N are midpoints of AB and CD respectively. Show that segments DM and BN trisect segment AC.
(ii) Use part (i) to find a procedure to trisect any given segment PQ. Find two other ways to trisect PQ, one using the fact that a line parallel to one side of a triangle divides the other two sides in the same ratio and a third using the Centroid Theorem.
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