Varignon parallelogram · 5 marks

Suppose PQRS is the Varignon parallelogram of ABCD.

(i) Copy only PQRS on another paper. Show how you will recreate a congruent copy A′B′C′D′ of ABCD from PQRS. This will be relevant when we return to study tilings later. (Hint: How will you place vertex A′? How will you place B′, C′ and D′?)
(ii) There are multiple ways to construct A′B′C′D′ in (i). Justify why the quadrilateral A′B′C′D′ you constructed is congruent to ABCD. You may need to show why S is collinear with the points A′ and D′ that you constructed, and similarly for P, Q and R. (Hint: Use congruence of triangles, for example ∆SDR ≅ ∆SD′R.)
(iii) Show that if PQRS is a square, then AC and BD are perpendicular and equal. Prove the converse.

Answer: (i) Place A′ so that ∆SA′P ≅ ∆SAP (copy SA and PA, same side as A). Extend A′P to B′ with PB′ = A′P, B′Q to C′ with QC′ = B′Q, C′R to D′ with RD′ = C′R; join D′A′. (ii) Step by step ∆PB′Q ≅ ∆PBQ, ∆QC′R ≅ ∆QCR, ∆RD′S ≅ ∆RDS (SAS); the angles at S add to 180°, so A′, S, D′ are collinear and the four corner triangles and PQRS match those of ABCD: A′B′C′D′ ≅ ABCD. (iii) PQ = ½AC, QR = ½BD, PQ ‖ AC, QR ‖ BD: PQRS a square ⇔ AC = BD and AC ⟂ BD.

Step-by-step solution

Idea: Each vertex of ABCD is the mirror image of its neighbour through the midpoint of the side between them (B is A “reflected” through P, and so on). So once one vertex A′ is placed correctly, the others follow by doubling segments through P, Q, R.

(i) Copy only PQRS on another paper. Show how you will recreate a congruent copy A′B′C′D′ of ABCD from PQRS. This will be relevant when we return to study tilings later. (Hint: How will you place vertex A′? How will you place B′, C′ and D′?)

Place A′: measure SA and PA on the original. On the copy, draw arcs of radius SA from S and PA from P; take their crossing point on the same side of SP as A. Then ∆SA′P ≅ ∆SAP.½ mark
Place B′, C′, D′: extend A′P beyond P to B′ with PB′ = A′P. Extend B′Q beyond Q to C′ with QC′ = B′Q. Extend C′R beyond R to D′ with RD′ = C′R. Join A′B′C′D′.1 mark

Copy ∆SAP to fix A′; then double A′P through P to get B′, double B′Q through Q to get C′, double C′R through R to get D′.

(ii) There are multiple ways to construct A′B′C′D′ in (i). Justify why the quadrilateral A′B′C′D′ you constructed is congruent to ABCD. You may need to show why S is collinear with the points A′ and D′ that you constructed, and similarly for P, Q and R. (Hint: Use congruence of triangles, for example ∆SDR ≅ ∆SD′R.)

∆SA′P ≅ ∆SAP (SSS, by construction), so ∠A′PS = ∠APS. Since A, P, B and A′, P, B′ are straight lines, ∠B′PQ = 180° − ∠A′PS − ∠SPQ = ∠BPQ. With PB′ = PA′ = PA = PB and PQ common, ∆PB′Q ≅ ∆PBQ (SAS), so QB′ = QB = QC.½ mark
In the same way ∆QC′R ≅ ∆QCR and then ∆RD′S ≅ ∆RDS (SAS each time), so SD′ = SD = SA = SA′ and ∠RSD′ = ∠RSD.½ mark
S is on A′D′: ∠A′SP + ∠PSR + ∠RSD′ = ∠ASP + ∠PSR + ∠RSD = 180°, because A, S, D are collinear. So A′, S, D′ are collinear and S is the midpoint of A′D′. (P, Q, R lie on A′B′, B′C′, C′D′ by construction.)½ mark
So A′B′C′D′ is made of PQRS and four corner triangles congruent to those of ABCD, placed the same way. Its sides are A′B′ = 2PA = AB, B′C′ = BC, C′D′ = CD, D′A′ = DA, and its angles match: A′B′C′D′ ≅ ABCD.½ mark

Each corner triangle of A′B′C′D′ is congruent to the matching corner triangle of ABCD (SSS, then SAS three times), and A′, S, D′ are collinear, so A′B′C′D′ ≅ ABCD.

(iii) Show that if PQRS is a square, then AC and BD are perpendicular and equal. Prove the converse.

Midpoint Theorem: PQ ‖ AC with PQ = ½AC (in ∆ABC), and QR ‖ BD with QR = ½BD (in ∆BCD).½ mark
If PQRS is a square: PQ = QR gives AC = BD; and PQ ⟂ QR gives AC ⟂ BD (lines parallel to two perpendicular lines are perpendicular).½ mark
Converse: if AC = BD and AC ⟂ BD, then PQ = QR and PQ ⟂ QR. A parallelogram with two equal adjacent sides meeting at a right angle is a square: PQRS is a square.½ mark

PQRS is a square exactly when AC = BD and AC ⟂ BD.

(i) Fix A′ by copying ∆SAP, then double through P, Q, R to get B′, C′, D′. (ii) Congruent corner triangles (SSS, SAS) and the collinearity of A′, S, D′ show A′B′C′D′ ≅ ABCD. (iii) PQRS is a square if and only if the diagonals AC and BD are equal and perpendicular.

Check: Why A′ must be placed carefully: if A′ is put anywhere, the doubling steps still close up (S is still the midpoint of D′A′) and still give a quadrilateral with Varignon parallelogram PQRS, but it is usually not congruent to ABCD. Many different quadrilaterals share the same PQRS.

Answer to write in the exam

(i)

Draw arcs: centre S radius SA, centre P radius PA; A′ = their meeting point on A’s side of SP

Produce A′P to B′ with PB′ = A′P

Produce B′Q to C′ with QC′ = B′Q

Produce C′R to D′ with RD′ = C′R; join A′B′C′D′

(ii)

∆SA′P ≅ ∆SAP (SSS, construction)

∠B′PQ = 180° − ∠A′PS − ∠SPQ = ∠BPQ; PB′ = PB, PQ common ⇒ ∆PB′Q ≅ ∆PBQ (SAS)

Similarly ∆QC′R ≅ ∆QCR, ∆RD′S ≅ ∆RDS (SAS)

∠A′SP + ∠PSR + ∠RSD′ = ∠ASP + ∠PSR + ∠RSD = 180° ⇒ A′, S, D′ collinear

∴ A′B′C′D′ = PQRS + four corner triangles congruent to those of ABCD ⇒ A′B′C′D′ ≅ ABCD

(iii)

PQ ‖ AC, PQ = ½AC; QR ‖ BD, QR = ½BD (Midpoint Theorem)

PQRS square ⇒ PQ = QR, PQ ⟂ QR ⇒ AC = BD, AC ⟂ BD

Conversely AC = BD, AC ⟂ BD ⇒ PQ = QR, ∠PQR = 90° ⇒ parallelogram PQRS is a square

Common mistakes that cost marks

Placing A′ anywhere. The doubling always closes up, but only the correct A′ gives a copy congruent to ABCD.
In (ii), assuming D′, S, A′ are collinear. This must be shown using the angles at S.
In (iii), proving only one direction. “Prove the converse” needs the reverse argument too.

How this can come in the exam

MCQ (1 mark)

The quadrilateral formed by joining the midpoints of the sides of quadrilateral ABCD is a square of side 5 cm. Then

AC = BD = 5 cm
AC = BD = 10 cm and AC ⟂ BD
AC = 10 cm, BD = 5 cm
AC ‖ BD

Show answer

(B) AC = BD = 10 cm and AC ⟂ BD
Each side of the midpoint figure is half a diagonal, and adjacent sides are parallel to AC and BD. A square of side 5 means AC = BD = 10 cm and AC ⟂ BD.

Try one yourself

P, Q, R, S are the midpoints of the sides of a square ABCD of side 8 cm. Show that PQRS is a square and find its side.