Midpoint theorem · 5 marks

(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.

(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.

Answer: (i) MBND is a parallelogram, so DM ‖ BN. With X = DM ∩ AC and Y = BN ∩ AC: in ∆ABY, MX ‖ BY through the midpoint M, so AX = XY; in ∆CDX, NY ‖ DX through the midpoint N, so CY = YX. Hence AX = XY = YC. (ii) Three ways: (a) build a parallelogram with diagonal PQ and use (i); (b) mark 3 equal steps on any ray from P, join the last to Q and draw parallels; (c) make PQ a median of a triangle: the centroid is ⅔ of the way from P.

Step-by-step solution

Idea: DM and BN are parallel lines through the midpoints of two sides. The converse of the Midpoint Theorem, used in two triangles, makes the three pieces of AC equal.

(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.

MB = ½AB = ½DC = DN and MB ‖ DN (parts of AB ‖ DC). So MBND is a parallelogram and DM ‖ BN.1 mark
Let DM meet AC at X and BN meet AC at Y. In ∆ABY, M is the midpoint of AB and MX ‖ BY, so X is the midpoint of AY: AX = XY.1 mark
In ∆CDX, N is the midpoint of CD and NY ‖ DX, so Y is the midpoint of CX: CY = YX. Hence AX = XY = YC: DM and BN trisect AC.1 mark

AX = XY = YC, so DM and BN trisect 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.

Way 1 (part (i)): choose any point B not on PQ. Complete parallelogram PBQD with PQ as a diagonal (draw through Q a line ‖ PB and through P a line ‖ BQ; they meet at D). Mark the midpoints M of PB and N of QD. Join DM and BN: they cut PQ into three equal parts.1 mark
Way 2 (equal parts and parallels): draw any ray from P and mark three equal steps P₁, P₂, P₃ with compasses. Join P₃Q and draw lines through P₁, P₂ parallel to P₃Q; they meet PQ at the trisection points (parallels cutting equal pieces from one side cut equal pieces from the other). Way 3 (centroid): choose a point B off line PQ and extend BQ to C with QC = BQ, so PQ is a median of ∆PBC. Join C to the midpoint of PB; it meets PQ at the centroid G with PG = ⅔PQ. The midpoint of PG gives the other trisection point.1 mark

Way 1: parallelogram with diagonal PQ and lines from its vertices to midpoints of sides. Way 2: three equal steps on a ray and parallel lines. Way 3: make PQ a median; the centroid is a trisection point and the midpoint of PG is the other.

(i) DM and BN cut AC into three equal parts. (ii) Trisect PQ by (a) a parallelogram on diagonal PQ, (b) equal steps on a ray with parallel lines, or (c) the centroid of a triangle with median PQ (plus the midpoint of PG).

Check: A(0, 0), B(2, 6), C(10, 6), D(8, 0): M(1, 3), N(9, 3). DM meets AC at (10/3, 2) and BN meets AC at (20/3, 4); these are ⅓ and ⅔ of the way along AC ✓.

Answer to write in the exam

(i)

MB = DN, MB ‖ DN ⇒ MBND is a parallelogram ⇒ DM ‖ BN

Let DM ∩ AC = X, BN ∩ AC = Y

∆ABY: M midpoint of AB, MX ‖ BY ⇒ AX = XY (converse of Midpoint Theorem)

∆CDX: N midpoint of CD, NY ‖ DX ⇒ CY = YX

∴ AX = XY = YC

(ii)

Way 1: parallelogram PBQD on diagonal PQ; M, N midpoints of PB, QD; DM, BN trisect PQ (part (i))

Way 2: ray from P, 3 equal steps P₁, P₂, P₃; join P₃Q; parallels to P₃Q through P₁, P₂ trisect PQ

Way 3: QC = BQ ⇒ PQ is a median of ∆PBC; median from C meets PQ at centroid G, PG = ⅔PQ; midpoint of PG is the other point

Common mistakes that cost marks

Assuming DM ‖ BN without proof; it follows from MBND being a parallelogram.
Using the converse of the Midpoint Theorem in ∆ABC, where X is not on a side.
In Way 3, stopping at the centroid: it gives only one trisection point; halve PG for the other.

How this can come in the exam

Short answer (2 marks)

In parallelogram ABCD, M and N are the midpoints of AB and CD, and DM, BN meet diagonal AC at X and Y. If AC = 15 cm, find XY and AY.

Show answer

DM and BN trisect AC (1 mark), so XY = 5 cm and AY = 10 cm (1 mark).

Try one yourself

Explain why, in Way 3, the centroid G of ∆PBC is a trisection point of PQ.