Varignon parallelogram · 4 marks

Suppose the midpoints of sides AB, BC, CD and DA of a quadrilateral ABCD are P, Q, R and S respectively.

(i) Show that PR and QS bisect each other.
(ii) Show that if AC = BD, then PR and QS are perpendicular. Is the converse true?

Answer: (i) PQRS is a parallelogram (midpoints of a quadrilateral), and PR, QS are its diagonals, so they bisect each other. (ii) If AC = BD then PQ = ½AC = ½BD = QR, so PQRS is a rhombus and its diagonals PR ⟂ QS. The converse is true: PR ⟂ QS makes the parallelogram PQRS a rhombus, so ½AC = ½BD, i.e. AC = BD.

Step-by-step solution

Idea: Everything follows from the parallelogram PQRS formed by the midpoints: its sides are half the diagonals of ABCD, PQ and SR parallel to AC, QR and PS parallel to BD.

(i) Show that PR and QS bisect each other.

By the Midpoint Theorem in ∆ABC and ∆ADC, PQ ‖ AC ‖ SR; in ∆BCD and ∆BAD, QR ‖ BD ‖ PS. So PQRS is a parallelogram.1 mark
PR and QS are the diagonals of parallelogram PQRS, and the diagonals of a parallelogram bisect each other.½ mark

PR and QS are the diagonals of the parallelogram PQRS, so they bisect each other.

(ii) Show that if AC = BD, then PR and QS are perpendicular. Is the converse true?

Midpoint Theorem: PQ = ½AC and QR = ½BD. If AC = BD then PQ = QR, so the parallelogram PQRS has two adjacent sides equal: it is a rhombus.1 mark
The diagonals of a rhombus are perpendicular, so PR ⟂ QS.½ mark
Converse (true): if PR ⟂ QS, the parallelogram PQRS has diagonals bisecting each other at right angles, so it is a rhombus. Then PQ = QR, i.e. ½AC = ½BD, so AC = BD.1 mark

PR ⟂ QS when AC = BD; and the converse is true: if PR ⟂ QS then AC = BD.

(i) PR and QS bisect each other (diagonals of the parallelogram PQRS). (ii) If AC = BD, PQRS is a rhombus, so PR ⟂ QS; conversely PR ⟂ QS forces AC = BD, so the converse is true.

Check: A(0, 0), B(6, 0), C(4, 3), D(3, 4): AC = 5 and BD = √(9 + 16) = 5. P(3, 0), Q(5, 1.5), R(3.5, 3.5), S(1.5, 2). PR = (0.5, 3.5) and QS = (−3.5, 0.5); (0.5)(−3.5) + (3.5)(0.5) = 0, so PR ⟂ QS ✓.

Answer to write in the exam

(i)

PQ ‖ AC ‖ SR and QR ‖ BD ‖ PS (Midpoint Theorem) ⇒ PQRS is a parallelogram

∴ Its diagonals PR and QS bisect each other.

(ii)

PQ = ½AC, QR = ½BD (Midpoint Theorem)

AC = BD ⇒ PQ = QR ⇒ parallelogram PQRS is a rhombus ⇒ PR ⟂ QS

Converse: PR ⟂ QS ⇒ PQRS (diagonals bisect at 90°) is a rhombus ⇒ PQ = QR ⇒ AC = BD

∴ The converse is true.

Common mistakes that cost marks

In (i), trying to prove the bisection with new congruent triangles. It is quicker to show PQRS is a parallelogram first.
In (ii), saying PQRS is a square. Equal diagonals of ABCD make PQRS a rhombus only.
Answering “the converse is false” without checking. The rhombus test works both ways here.

How this can come in the exam

Assertion–Reason (1 mark)

Assertion (A): The quadrilateral formed by joining the midpoints of the sides of an isosceles trapezium is a rhombus.
Reason (R): The diagonals of an isosceles trapezium are equal.

Both A and R are true, and R is the correct explanation of A.
Both A and R are true, but R is not the correct explanation of A.
A is true but R is false.
A is false but R is true.

Show answer

(A) Both A and R are true, and R is the correct explanation of A.
Equal diagonals make the sides of the midpoint parallelogram equal (each is half a diagonal), so it is a rhombus. R explains A.

Try one yourself

The midpoints of the sides of quadrilateral ABCD form a rectangle. What can you say about AC and BD?