Varignon parallelogram · 4 marks

Is the Midpoint Theorem for Quadrilaterals true when the quadrilateral is non-convex? How about when it is self-intersecting? Experiment and check. The pictures may look strange, but the theorem appears to be true.

(i) Can you explain why this is so? See the geometric reasoning in the text.
(ii) There is one exception: In a very special case the Varignon parallelogram becomes a single segment. When will this happen?
(iii) Does your reasoning apply even when ABCD is non-planar?

Answer: Yes, in both cases the midpoints form a parallelogram. (i) The proof uses only the Midpoint Theorem in the four triangles ABC, ADC, BCD, BAD, which exist however ABCD is shaped: PQ ‖ AC ‖ SR and QR ‖ BD ‖ PS. (ii) It becomes a segment exactly when AC ‖ BD (possible only for a self-intersecting ABCD), because then all four sides of PQRS lie along one direction. (iii) Yes: the four triangles still exist, PQ and SR are both parallel to AC, so P, Q, R, S lie in one plane and PQRS is a parallelogram.

Step-by-step solution

Idea: The Varignon argument never used convexity: each side of PQRS joins midpoints of two sides of a triangle with a diagonal of ABCD as third side.

(i) Can you explain why this is so? See the geometric reasoning in the text.

For any four points A, B, C, D (no three collinear), triangles ABC, ADC, BCD and BAD exist. Midpoint Theorem: PQ ‖ AC and PQ = ½AC (∆ABC); SR ‖ AC and SR = ½AC (∆ADC).1 mark
So PQ and SR are equal and parallel, and PQRS is a parallelogram. Nothing used whether ABCD is convex, non-convex or self-intersecting.½ mark

The proof only uses the Midpoint Theorem in triangles formed by three of the four points, which exist in every case.

(ii) There is one exception: In a very special case the Varignon parallelogram becomes a single segment. When will this happen?

The sides of PQRS are parallel to AC (PQ, SR) and to BD (QR, PS). If the lines AC and BD are parallel, all four sides lie along one direction, and PQRS flattens into a segment.1 mark
AC ‖ BD cannot happen in a convex quadrilateral (its diagonals cross) or a non-convex one, but it can in a self-intersecting one: e.g. A(0, 0), B(0, 2), C(4, 0), D(4, 2) gives P(0, 1), Q(2, 1), R(4, 1), S(2, 1): Q = S and PQRS is the segment PR.½ mark

When the diagonals AC and BD are parallel (a self-intersecting ABCD); then P, Q, R, S are collinear.

(iii) Does your reasoning apply even when ABCD is non-planar?

Yes. Each of the triangles ABC, ADC is still a flat triangle, so PQ ‖ AC ‖ SR and PQ = SR = ½AC. Two parallel lines lie in one plane, so P, Q, R, S are coplanar, and PQRS is a parallelogram (one pair of opposite sides equal and parallel).1 mark

Yes: PQ and SR are both parallel and equal to ½AC, so PQRS is a (flat) parallelogram even when ABCD is not.

Yes, the midpoints always form a parallelogram (also for non-convex, self-intersecting and non-planar ABCD), because each side of PQRS is parallel and equal to half a diagonal. It degenerates to a segment exactly when AC ‖ BD.

Check: Non-planar example: A(0, 0, 0), B(2, 0, 0), C(2, 2, 0), D(0, 2, 2). P(1, 0, 0), Q(2, 1, 0), R(1, 2, 1), S(0, 1, 1). Q − P = (1, 1, 0) = R − S ✓, so PQRS is a parallelogram.

Answer to write in the exam

(i)

∆ABC: PQ ‖ AC, PQ = ½AC; ∆ADC: SR ‖ AC, SR = ½AC (Midpoint Theorem)

⇒ PQ = SR, PQ ‖ SR ⇒ PQRS is a parallelogram

∴ True for non-convex and self-intersecting ABCD (convexity never used)

(ii)

PQ ‖ SR ‖ AC and QR ‖ PS ‖ BD

If AC ‖ BD, all sides of PQRS have one direction ⇒ PQRS is a segment

E.g. A(0, 0), B(0, 2), C(4, 0), D(4, 2): P(0, 1), Q = S = (2, 1), R(4, 1)

∴ Happens exactly when AC ‖ BD (only for a self-intersecting ABCD)

(iii)

∆ABC, ∆ADC are plane triangles ⇒ PQ ‖ AC ‖ SR, PQ = SR = ½AC

Parallel lines PQ, SR lie in one plane ⇒ P, Q, R, S coplanar

∴ PQRS is a parallelogram even for non-planar ABCD

Common mistakes that cost marks

Saying the theorem fails for a self-intersecting quadrilateral because the picture looks odd. The proof still works.
In (ii), answering “when ABCD is a parallelogram”. Then PQRS is still a parallelogram; the collapse needs AC ‖ BD.
In (iii), forgetting to explain why P, Q, R, S lie in one plane.

How this can come in the exam

MCQ (1 mark)

For any quadrilateral ABCD, the side PQ of its midpoint parallelogram (P on AB, Q on BC) is always

parallel to BD and half of it
parallel to AC and half of it
equal to AB
perpendicular to AC

Show answer

(B) parallel to AC and half of it
In ∆ABC, P and Q are midpoints of AB and BC, so PQ ‖ AC and PQ = ½AC.

Try one yourself

A(0, 0), B(6, 2), C(2, 2), D(1, 6) form a non-convex quadrilateral. Find its midpoint quadrilateral and check it is a parallelogram.