066i - Proofs with Quadrilateral Properties

Lesson 66 — Problem-Solving: Proofs Involving Quadrilateral Properties

Strand: Space | Descriptor: AC9M8SP02 | Duration: 45 minutes

Learning Intentions

• To construct formal proofs of quadrilateral properties using congruent triangles and angle reasoning.
• To reason in both directions: from a named shape to its properties, and from given properties to identifying the shape.

Success Criteria

I can:

1. Prove that the diagonals of a rhombus are perpendicular.
2. Prove that a parallelogram with one right angle must be a rectangle.
3. Prove the converse: that a quadrilateral whose diagonals bisect each other must be a parallelogram.
4. Justify every line of a proof with a specific reason.

Warmup

(5 minutes — true/false with justification, pairs)

1. The diagonals of a parallelogram always bisect each other.
2. The diagonals of a rhombus are always equal in length.
3. Opposite angles of a kite are always equal.
4. A rectangle’s diagonals always bisect the angles at each vertex.

Answers: 1. True (Lesson 63); 2. False — they bisect each other at right angles but are equal only in the special case of a square; 3. False — only the pair of angles between the equal sides is guaranteed equal; 4. False — only true for a square; a non-square rectangle’s diagonals do not bisect its angles into two equal parts unless it is a square.

Activities

Activity 1 — Explicit Instruction: a Formal Proof (12 min)

I do: Prove that the diagonals of a rhombus meet at right angles.

Statement	Reason
,	Diagonals of a parallelogram bisect each other (rhombus is a parallelogram)
	All sides of a rhombus are equal
	Common side
	SSS
	Corresponding angles in congruent triangles
	Angles on a straight line
	Combining the two lines above

Activity 2 — Guided Proof: Parallelogram with a Right Angle (10 min)

We do: Prove together that a parallelogram with must have all four angles equal to (i.e. it is a rectangle).

Statement	Reason
	Given
	Co-interior angles,
	Substitution
	Opposite angles of a parallelogram are equal
	Opposite angles of a parallelogram are equal

Prompt for discussion: why is it enough to know one angle of a parallelogram is to guarantee it’s a rectangle?

Activity 3 — Problem-solving: Proving the Converse (12 min)

Pairs, with scaffolding.

A quadrilateral has diagonals that bisect each other at (that is, and ). Prove that is a parallelogram.

Socratic scaffolding (Polya’s cycle):

Prompt	Purpose
Understand: what is given, and what must you show?	Given: diagonals bisect each other. To prove: both pairs of opposite sides are parallel.
What is different from Lesson 63’s proof?	Lesson 63 proved bisection from the parallelogram; here we must prove the parallelogram from the bisection — the reverse direction.
Devise a plan — can you find congruent triangles first?	Look at and . What is already equal?
Carry out the plan — which condition applies?	, , (vertically opposite) — SAS.
What does the congruence give you?	(corresponding angles), and .
How does this prove ?	Equal alternate angles on transversal prove the lines are parallel (this is exactly Lesson 61’s Activity 3 result).
Looking back — have you proven both pairs parallel?	Repeat the identical argument with to get . Only then is fully proven a parallelogram.

Checks for Understanding

(6 minutes — exit ticket, collected)

1. State the reason used to prove in the rhombus diagonal proof.
2. A parallelogram has . What can you immediately conclude about , and ?
3. Why is SAS, not SSS, the condition used in the converse proof (diagonals bisecting parallelogram)?
4. Reasoning. A quadrilateral has diagonals that bisect each other and are equal in length. What extra shape must it be, beyond a parallelogram? Justify briefly.

Answers: 1. Angles on a straight line sum to , combined with the two angles being equal (from the congruent triangles); 2. All four angles must be — it is a rectangle; 3. Only two sides (, ) and one angle (vertically opposite) are known at the start — there is no third side known yet, so SSS is not available; 4. A rectangle — equal, bisecting diagonals is the defining congruent-triangle setup (SSS once the diagonal lengths are equal) that proves all four angles are .

Common Misconceptions

Misconception	How to pre-empt it
Proving one direction (shape property) is treated as automatically proving the converse (property shape).	Explicitly label each proof “forward” or “converse” and discuss why both directions need separate justification.
Skipping the second pair of sides when proving a quadrilateral is a parallelogram from diagonal bisection.	Require students to explicitly repeat the argument for the second diagonal pair, not just assume symmetry.
Believing “diagonals bisect each other” alone proves a rhombus.	Contrast with a non-rhombus parallelogram (e.g. a “stretched” rectangle) where diagonals bisect but are unequal and not perpendicular.
Writing “opposite angles equal” as the reason for a rectangle’s right angles, without linking back to the given right angle.	Insist the co-interior angle step is shown explicitly, not skipped.
Circular reasoning — using the property being proved as a reason partway through its own proof.	Have students check off which facts are “given,” “previously proved,” or “still to prove” before writing each line.

Enrichment — Competition-Style Problems

E1 (Challenge proof). Prove that if a parallelogram has diagonals that bisect each other at right angles, it must be a rhombus.

Answer

Let diagonals meet at with , , and . In and : , (common), — SAS, so , giving . Combined with opposite sides already equal (parallelogram property), all four sides are equal — a rhombus.

E2 (AMC Junior style). is a rectangle. is the midpoint of diagonal . Explain why is also the midpoint of diagonal .

Answer

A rectangle is a parallelogram, and the diagonals of any parallelogram bisect each other. Since is defined as the midpoint of and the diagonals bisect at the same point, must also be the midpoint of .

E3 (Investigation — Varignon’s theorem). The midpoints of the sides of any quadrilateral (even an irregular one) are joined in order. Investigate: what shape is always formed?

Answer

A parallelogram — always, regardless of the original quadrilateral’s shape. Each side of the midpoint quadrilateral is parallel to, and half the length of, a diagonal of the original quadrilateral (provable using similar triangles on the diagonal), so opposite sides of the midpoint quadrilateral are both parallel and equal.

Homework

1. Write the full statement–reason proof that the diagonals of a rhombus bisect its vertex angles (use , SSS).
2. A parallelogram has . State, with reasons, the size of the other three angles.
3. Explain in your own words why proving “shape property” does not automatically prove “property shape.”
4. A quadrilateral has diagonals that bisect each other. Name the congruence condition used to begin a proof that it is a parallelogram.
5. Reasoning. A square is a rhombus, a rectangle, and a parallelogram all at once. Explain how this “nesting” of proofs means every rectangle-property proof and every rhombus-property proof automatically applies to a square.
6. Challenge. Prove that if a quadrilateral has one pair of opposite sides that are both equal and parallel, it must be a parallelogram. (Hint: draw one diagonal and look for congruent triangles.)

Answers: 1. Using diagonal : , (rhombus sides equal), common — SSS, so , giving and , i.e. the diagonal bisects the angles at and ; 2. All three other angles are (co-interior and opposite-angle-equal reasoning) — it is a rectangle; 3. A forward proof only shows the property is a consequence of the shape; it says nothing about whether other shapes could also produce that property, so the reverse implication needs its own independent proof; 4. SAS (using the bisected diagonals and vertically opposite angles at the intersection point); 5. Since a square satisfies every defining property of a rhombus and of a rectangle, any theorem proved “for all rhombi” or “for all rectangles” applies automatically to squares as a special case — no new proof is needed; 6. Let and in quadrilateral . Draw diagonal . Since , (alternate angles, transversal ). With and common, by SAS, giving , which are alternate angles proving — so both pairs of opposite sides are parallel, confirming a parallelogram.