Proving a parallelogram is a rectangle or a rhombus

When you reach for it

A proof or a problem already has a parallelogram, or can get one from the tests for a parallelogram, and now needs it to be something more special: a rectangle, a rhombus, or both at once, a square. The targets are four right angles or four congruent sides, and neither is easy to reach directly. A proof rarely hands over four angles or four lengths. What it usually supplies is one right angle, a pair of equal diagonals, one pair of equal sides, a right angle where the diagonals cross, or a diagonal that cuts a corner in half. A parallelogram already carries a lot of structure, and this lesson finds how little has to be added to it for each target.

The words, exactly. As in rectangles, rhombuses and squares:

Rectangle. A rectangle is a parallelogram with four right angles.

Rhombus. A rhombus is a parallelogram whose four sides are all congruent.

Square. A square is a parallelogram that is both a rectangle and a rhombus: four right angles and four congruent sides.

These definitions are inclusive. The rectangle definition asks about angles and says nothing about sides, so a square, with four right angles, is a rectangle. The rhombus definition asks about sides and says nothing about angles, so a square is a rhombus too, and every rhombus test in this lesson is passed by squares as well as every rectangle test.

The lesson axioms, definitions and theorems states both definitions without the word “parallelogram”: a rectangle as a quadrilateral with four right angles, and a rhombus as a quadrilateral whose four sides all have the same length. Each pair of wordings picks out exactly the same shapes, so either can be used. One way round is immediate in both cases: a parallelogram with four right angles is a quadrilateral with four right angles, and the same for four congruent sides. The other way round is a theorem in each case, and each is one of the tests below. A quadrilateral with four right angles is a parallelogram by the argument of the Three Right Angles Test. A quadrilateral ABCD with four congruent sides has \overline{AB} \cong \overline{CD} and \overline{BC} \cong \overline{DA}, both pairs of opposite sides congruent, so it is a parallelogram by the Opposite Sides Converse in proving a quadrilateral is a parallelogram; that is the Four Sides Test. Here the parallelogram wordings are the ones in force, because they say what is already known and what is still missing.

Names. Throughout, ABCD is a quadrilateral named in order around it, as in proving a quadrilateral is a parallelogram: four segments \overline{AB}, \overline{BC}, \overline{CD} and \overline{DA} that meet only at their ends, with no two neighboring sides along one line, convex or not. So \overline{AB} and \overline{BC} are neighbors, meeting at B, while \overline{AB} is opposite \overline{CD}, and the diagonals are \overline{AC} and \overline{BD}. The angle \angle A means the angle of the quadrilateral at A, which is \angle DAB.

What a parallelogram already gives. If ABCD is a parallelogram, then, from properties of a parallelogram:

• opposite sides are congruent (proving properties of a parallelogram);
• opposite angles are congruent, and neighboring angles are same-side interior angles between parallel sides, so they add to 180^\circ;
• the diagonals bisect each other (proving the diagonals bisect each other).

None of these says anything about a right angle or about two neighboring sides, since a leaning parallelogram with one long pair of sides and one short pair has all of them. So one more fact is needed, and the tests below say which facts are enough.

Tests for a rectangle.

Tests for a rhombus.

The tests that start from a parallelogram are false without it; the Boundaries show a shape for each. The tests that start from any quadrilateral have hypotheses strong enough to build the parallelogram first.

The procedure

Proving a parallelogram is a rectangle

The target is the definition. Every proof below ends by showing that a parallelogram has four right angles. Since the parallelogram hands over its angles in linked pairs, one right angle spreads to the other three, and most of the work is getting that first one.

The One Right Angle Test. If a parallelogram has one right angle, it is a rectangle.

Given. ABCD is a parallelogram, and m\angle A = 90^\circ.

Prove. ABCD is a rectangle.

Line 3 needs \angle DAB and \angle CBA to lie on the same side of \overline{AB}, between the parallel lines. They do: \overline{DC} is parallel to \overline{AB}, so it never meets \overleftrightarrow{AB}, and its two ends C and D are on the same side of it.

Any corner can play the part of \angle A, so a right angle anywhere is enough. The four angles of a parallelogram are only ever two sizes, x and 180^\circ - x, and a right angle is the one size that is its own supplement.

The Equal Neighbors Test. If two neighboring angles of a parallelogram are congruent, it is a rectangle. Neighboring angles, such as \angle A and \angle B, add to 180^\circ, as in line 3 above. If they are also equal, each is half of 180^\circ:

and the One Right Angle Test finishes. The next proof ends with this test.

The Equal Diagonals Test. If the diagonals of a parallelogram are congruent, it is a rectangle.

Given. ABCD is a parallelogram, and \overline{AC} \cong \overline{BD}.

Prove. ABCD is a rectangle.

The Given is about lengths and the target is about an angle, and congruent triangles are how lengths become angles. The two triangles that stand on the side \overline{AB}, one using each diagonal, have every side known: the diagonals are given equal, the sides \overline{AD} and \overline{BC} are equal because they are opposite, and \overline{AB} is shared.

In line 4 the correspondence is D \to C, A \to B, B \to A (congruence statements and correspondence). Under it \overline{DA} matches \overline{CB} (line 2), \overline{AB} matches \overline{BA} (line 3), and \overline{DB} matches \overline{CA} (line 1). The angle at A in \triangle DAB then matches the angle at B in \triangle CBA, which is line 5. Those are the corner angles \angle A and \angle B of the parallelogram, neighbors along \overline{AB}, which is what line 6 needs.

The other direction. A rectangle’s diagonals are congruent, and the proof uses the same two triangles. In a rectangle \overline{AD} \cong \overline{BC} (opposite sides), \angle DAB \cong \angle CBA (both right) and \overline{AB} \cong \overline{BA}, so \triangle DAB \cong \triangle CBA by SAS, and \overline{DB} \cong \overline{CA}. Together the two directions make one “if and only if” statement: a parallelogram is a rectangle if and only if its diagonals are congruent. Each direction needed its own proof, SAS one way and SSS the other (a theorem and its converse).

Starting from any quadrilateral. Sometimes nothing says the shape is a parallelogram. Two tests still work, because each hypothesis builds the parallelogram on the way.

The Three Right Angles Test. If a quadrilateral has three right angles, it is a rectangle. The four angles add to 360^\circ (angles in a quadrilateral), so the fourth is 360^\circ - 3 \cdot 90^\circ = 90^\circ. Every angle is then less than 180^\circ, so the quadrilateral is convex, and C and D lie on the same side of \overleftrightarrow{AB}. So \angle A and \angle B are same-side interior angles for the lines \overleftrightarrow{AD} and \overleftrightarrow{BC}, and they add to 180^\circ, so \overline{AD} \parallel \overline{BC} by the converse of the Same-Side Interior Angles Theorem. In the same way \angle A and \angle D give \overline{AB} \parallel \overline{DC}. So ABCD is a parallelogram with four right angles. The same argument, run on four right angles, is why the two wordings of the definition of a rectangle agree.

The Equal Halves Test. If the diagonals of a quadrilateral bisect each other and are congruent, it is a rectangle. Diagonals that bisect each other make a parallelogram, by the Diagonals Converse in proving a quadrilateral is a parallelogram, and the Equal Diagonals Test finishes. If the diagonals meet at E, the two conditions together say exactly that E is the same distance from all four corners:

since two equal diagonals, each cut in half, have four equal halves; and conversely four equal halves make each diagonal twice the same length.

Proving a parallelogram is a rhombus

The target is two congruent neighbors. A parallelogram already has its opposite sides congruent, so it is two lengths away from a rhombus, not four. Each proof below ends by producing one pair of congruent neighboring sides, and the first two tests turn that pair into four.

The Four Sides Test. If all four sides of a quadrilateral are congruent, the quadrilateral is a rhombus.

This is the argument given under The words, exactly. The Opposite Sides Converse was proved for every quadrilateral, convex or not, so this test needs nothing about the shape beyond its four lengths.

The Neighboring Sides Test. If two neighboring sides of a parallelogram are congruent, the parallelogram is a rhombus.

Given. ABCD is a parallelogram and \overline{AB} \cong \overline{BC}.

The Opposite Sides Theorem gives \overline{CD} \cong \overline{AB} and \overline{DA} \cong \overline{BC}. With the Given and the transitive property of congruence, all four sides are congruent to \overline{AB}, so ABCD is a rhombus by definition. This short argument was first met as a worked example in proving properties of a parallelogram; here it becomes a test that later proofs can cite in one line. The two remaining tests both end with it.

The Perpendicular Diagonals Test. If the diagonals of a parallelogram are perpendicular, the parallelogram is a rhombus.

Given. ABCD is a parallelogram and \overline{AC} \perp \overline{BD}.

Prove. ABCD is a rhombus.

A right angle and a shared segment point to congruent triangles, and the parallelogram supplies the last piece: its diagonals cut each other in half.

In line 6 the right angle at E lies between \overline{EA} and \overline{EB} in one triangle and between \overline{EC} and \overline{EB} in the other, so the correspondence is A \to C, E \to E, B \to B, and \overline{AB} matches \overline{CB}. Those two sides meet at B, so they are neighbors, which is exactly what line 8 needs.

The parallelogram is used twice: in line 8, and in line 2, which takes only half of what the diagonals theorem offers, that E is the midpoint of \overline{AC}. The other half, \overline{BE} \cong \overline{DE}, is never needed. The Boundaries show why that matters.

The Angle Bisector Test. If a diagonal of a parallelogram bisects the angle at one of its ends, the parallelogram is a rhombus.

Given. ABCD is a parallelogram and \overline{AC} bisects \angle DAB, so \angle BAC \cong \angle DAC.

Prove. ABCD is a rhombus.

The parallel sides move one half of the angle at A across to C. Then \triangle ACD has two congruent angles, and an isosceles triangle hands over two congruent sides.

In line 5 the two congruent angles of \triangle ACD sit at A and at C. The side opposite the angle at A is \overline{DC} and the side opposite the angle at C is \overline{DA}, so those are the congruent sides, and they meet at D.

Line 3 needs B and D on opposite sides of \overleftrightarrow{AC}, so that \angle BAC and \angle DCA really are alternate interior angles. In a parallelogram that always holds: the diagonals cross at their shared midpoint, which lies on \overleftrightarrow{AC} and strictly between B and D.

The hypothesis is small: one diagonal, and the angle at only one of its two ends. Which diagonal and which end does not matter, since the letters can be renamed. And once ABCD is a rhombus, the same diagonal bisects the angle at its other end as well (see below).

Both directions. Each rhombus test is the converse of a property of a rhombus, and a converse needs its own proof (a theorem and its converse). The forward properties are short to prove, so here they are, which makes every condition in the rhombus table an “if and only if” for parallelograms.

Let ABCD be a rhombus with diagonals meeting at E.

• Each diagonal bisects both angles at its ends. In \triangle ABC and \triangle ADC, \overline{AB} \cong \overline{AD} and \overline{CB} \cong \overline{CD} because all four sides are congruent, and \overline{AC} is shared, so the triangles are congruent by SSS with B \to D. Corresponding parts give \angle BAC \cong \angle DAC and \angle BCA \cong \angle DCA: \overline{AC} bisects \angle A and \angle C. The same argument with \triangle BAD and \triangle BCD shows that \overline{BD} bisects \angle B and \angle D.
• The diagonals are perpendicular. A rhombus is a parallelogram, so E is the midpoint of \overline{BD}. In \triangle AEB and \triangle AED, \overline{AB} \cong \overline{AD}, \overline{BE} \cong \overline{DE} and \overline{AE} is shared, so they are congruent by SSS and \angle AEB \cong \angle AED. These two angles form a linear pair, so they add to 180^\circ, and being equal, each is 90^\circ.

Put together with the tests, this gives three “if and only if” statements. For a parallelogram ABCD, each of these holds exactly when ABCD is a rhombus: two neighboring sides are congruent, such as \overline{AB} \cong \overline{BC}; the diagonals are perpendicular, \overline{AC} \perp \overline{BD}; a diagonal bisects the angle at one of its ends, such as \overline{AC} bisecting \angle A.

The two directions do different jobs in a problem, for a rectangle as for a rhombus. A test (the converse) says a condition is enough: if it holds, the figure is a rhombus. A property (the forward direction) says the condition is needed: if it fails, the figure is not a rhombus. Solving for an unknown that “makes ABCD a rhombus” uses both, as Example 5 shows, and Example 2 does the same for a rectangle.

Both at once: proving a square

A square is a parallelogram that is a rectangle and a rhombus, so a square is proved in two halves: one test from each table, each applied with its own starting point.

Starting from a parallelogram. A parallelogram whose diagonals are congruent and perpendicular is a square: the Equal Diagonals Test makes it a rectangle and the Perpendicular Diagonals Test makes it a rhombus. So is a parallelogram with one right angle and two congruent neighboring sides, by the One Right Angle Test and the Neighboring Sides Test. Any rectangle test paired with any rhombus test works the same way.

Starting from any quadrilateral. The tests that build the parallelogram for themselves pair up too. Diagonals that bisect each other, are congruent and are perpendicular make a square: the Equal Halves Test gives a rectangle, which is a parallelogram, and the Perpendicular Diagonals Test then gives a rhombus. Four congruent sides and one right angle make a square: the Four Sides Test gives a rhombus, which is a parallelogram, and the One Right Angle Test then gives a rectangle. Without a parallelogram from somewhere, the pairing fails: congruent diagonals that cross at right angles are not enough on their own (Exercise 21).

Neither kind of test alone can decide “square”: the rectangle tests say nothing about the sides, and the rhombus tests say nothing about the angles. In the other direction, a square is a rectangle and a rhombus, so it has every property of both: its diagonals are congruent, perpendicular, bisect each other and bisect the corner angles.

Choosing a test

Let the information decide, and first decide which target it serves. Facts about angles at the corners, and about the lengths of the diagonals, lead to a rectangle. Facts about sides, about the angle at which the diagonals cross, and about angles along a diagonal lead to a rhombus.

For a rectangle. A right angle that is given, or that comes from perpendicular lines, points to the One Right Angle Test. Lengths of diagonals point to the Equal Diagonals Test, and so do facts about the point where they cross, once the halves are doubled. Angles found equal from congruent triangles, especially two at the ends of one side, point to the Equal Neighbors Test.

For a rhombus. Lengths of sides point to the Four Sides Test, or to the Neighboring Sides Test once the figure is known to be a parallelogram. A right angle where the diagonals cross points to the Perpendicular Diagonals Test, and angles along a diagonal point to the Angle Bisector Test.

When the parallelogram is not known yet. Use a test that starts from any quadrilateral (Three Right Angles, Equal Halves, Four Sides), or first prove the shape is a parallelogram with a test from proving a quadrilateral is a parallelogram.

In coordinates. The shortest route is usually two midpoints that agree, which makes a parallelogram, and then a calculation. For a rectangle: two diagonal lengths from the distance formula, or two sides with slopes whose product is -1 (perpendicular slopes). For a rhombus: two diagonals with slopes whose product is -1, or two neighboring sides of equal length; or, skipping the midpoints, four equal side lengths.

Worked examples

Boundaries

The rectangle tests need the parallelogram

Each of the first three rectangle tests begins “if a parallelogram”, and without that word each one fails.

• One right angle, or two. A quadrilateral can have right angles at two neighboring corners and not be a rectangle, as in Example 1(e), or at two opposite corners, as in the kite below. Its other two angles are 60^\circ and 120^\circ. Three right angles are enough, since the angle sum then forces the fourth; two never are.

• Equal diagonals. An isosceles trapezoid, with congruent legs that are not parallel, has congruent diagonals and is not a parallelogram, as proved in proving the isosceles trapezoid properties. A quadrilateral need not even be a trapezoid: the one below has congruent diagonals and no two sides parallel. What it lacks is diagonals that bisect each other, which is the other half of the Equal Halves Test.

• Equal neighboring angles. In an isosceles trapezoid the two angles at one base are neighbors and congruent, yet they are not right angles: if they were, both legs would be perpendicular to that base, and so parallel to each other. The Equal Neighbors Test works because in a parallelogram every pair of neighbors adds to 180^\circ, and a trapezoid’s base angles need not.

The rhombus tests need the parallelogram

Three rhombus tests need the word “parallelogram”. The kite below has all three of their conditions and is not a rhombus.

• Perpendicular diagonals. Its diagonals cross at right angles at E.
• A diagonal bisecting an angle at its end. \overline{AC} bisects \angle A, and by the symmetry of the kite it bisects \angle C too.
• Two congruent neighboring sides. \overline{AB} \cong \overline{AD}, and \overline{CB} \cong \overline{CD} as well.

Yet AB \ne CB. The proof of the Perpendicular Diagonals Test shows where it slips through. Line 2 needs E to be the midpoint of \overline{AC}, and in the kite it is not: E is the midpoint of \overline{BD} only. Running the proof with that half instead gives \triangle AEB \cong \triangle AED and so \overline{AB} \cong \overline{AD}, which is one more pair of congruent neighbors, and the Neighboring Sides Test, which would finish the job, needs a parallelogram.

Perpendicular diagonals need not even give a kite. The quadrilateral with vertices A(0, 4), B(-2, 0), C(0, -1) and D(3, 0) has its diagonals along the two axes, so they are perpendicular, yet its sides are \displaystyle \sqrt{20}, \displaystyle \sqrt{5}, \displaystyle \sqrt{10} and 5, no two equal.

Diagonals that are perpendicular bisectors of each other are enough. For a quadrilateral not known to be a parallelogram, both diagonal conditions must be there (Exercise 45). If only one diagonal is the perpendicular bisector of the other, the shape is a kite. This is the rhombus partner of the Equal Halves Test, where diagonals that bisect each other and are congruent make a rectangle.

Three congruent sides are not enough. A(3, 4), B(8, 4), C(11, 0) and D(0, 0) give AB = 5, \displaystyle BC = \sqrt{3^2 + 4^2} = 5 and \displaystyle DA = \sqrt{3^2 + 4^2} = 5, but CD = 11. This is an isosceles trapezoid. The Four Sides Test needs the fourth side too.

Each kind of test decides only its own kind

A rectangle test says nothing about the sides. A rectangle need not be a square, as in Example 3, where the neighboring sides are \displaystyle 3\sqrt{5} and \displaystyle 2\sqrt{5}.

A rhombus test says nothing about the angles. A rhombus need not be a square either, as in Example 6, where \angle B is not a right angle.

Congruent diagonals belong to the rectangle, perpendicular ones to the rhombus. A rectangle with sides 4 and 3 has congruent diagonals and is no rhombus, and a rhombus with angles of 60^\circ and 120^\circ has perpendicular diagonals and is no rectangle. A rhombus has congruent diagonals only when it is also a square, and a rectangle has perpendicular diagonals only when it is also a square.

Inclusive definitions. A square passes every test in this lesson, and under the inclusive definitions used here that is correct, because a square is a rectangle and a rhombus. A book that defined a rectangle or a rhombus to exclude squares would find these tests proving only “a rectangle or a square” or “a rhombus or a square”, and would need a further condition to rule the square out.

Shapes that are not convex

No shape that is not convex is a rhombus, or a rectangle. A dart, a kite pushed in, can have two pairs of congruent neighboring sides, a diagonal that bisects two of its angles, and diagonal lines at right angles, but it can never have four congruent sides (Exercise 54). And it can never have three right angles, since the Three Right Angles Test showed that three right angles force a convex shape.

The flat plane

The One Right Angle Test uses the Same-Side Interior Angles Theorem, the Angle Bisector Test uses the Alternate Interior Angles Theorem, and the Three Right Angles Test uses the angle sum of a quadrilateral. All three rest on the parallel postulate, so the tests here are statements about the ordinary flat plane.

Where this goes next

The rectangle and the rhombus are the two branches of the family below the parallelogram: one reached by a fact about angles or the lengths of the diagonals, the other by a fact about sides or the angle between the diagonals. A shape on both branches is a square, and how all of them sit together, with the kite and the trapezoid beside them, is the quadrilateral hierarchy. A rhombus is exactly a kite whose two pairs of equal sides are equal to each other (trapezoids and kites).

Perpendicular diagonals are also what give a rhombus its area formula. They cut it into four right triangles, and the area comes out as half the product of the diagonals, in the area of a rhombus or kite.

On a grid the tests become two midpoints and then two lengths or two slopes, as in Examples 3, 6 and 7, which is how a shape is named in classifying a quadrilateral on the plane and how general facts are proved in coordinate proofs.

Example 8 is the start of a link with circles, and it runs both ways. A rectangle’s diagonals are congruent and bisect each other, so the point where they cross is the same distance from all four corners, and every rectangle has a circle through its corners, with its diagonals as diameters. A quadrilateral with that property is a cyclic quadrilateral, and its opposite angles add to 180^\circ, as proved in proving the cyclic quadrilateral angles. For a parallelogram, whose opposite angles are also equal, that means 90^\circ each: a parallelogram that fits in a circle is a rectangle.

And Example 10 is one case of a wider habit: a compass construction works because it builds a shape whose properties are already proved, here a rhombus and its perpendicular diagonals.

Exercises

In every exercise, ABCD is a quadrilateral named in order around it, with diagonals meeting at E where they cross, unless the exercise says otherwise.

Which facts make a rectangle

For each of 1 to 8, decide whether the facts guarantee that ABCD is a rectangle. If they do, name the test. If not, name a shape that has the facts and is not a rectangle, or explain why ABCD cannot be one.

1. ABCD is a parallelogram and m\angle A + m\angle C = 180^\circ.
2. ABCD is a parallelogram and AC = BD = 15.
3. ABCD is a parallelogram and m\angle A = 2 \cdot m\angle B.
4. ABCD is a parallelogram and \angle B \cong \angle D.
5. AC = BD, and nothing else is known.
6. \angle A, \angle B and \angle C are right angles.
7. The diagonals meet at E, with AE = BE = CE = DE = 5.
8. \angle B and \angle D are right angles.

Which facts make a rhombus

For each of 9 to 16, decide whether the facts guarantee that ABCD is a rhombus. If they do, name the test. If not, name a shape that has the facts and is not a rhombus.

9. ABCD is a parallelogram with AB = 9 and BC = 9.
10. ABCD is a parallelogram and m\angle BEC = 90^\circ.
11. \overline{AC} \perp \overline{BD}, and nothing else is known.
12. ABCD is a parallelogram and AC = BD.
13. ABCD is a parallelogram, m\angle ABD = 40^\circ and m\angle CBD = 40^\circ.
14. AB = BC = CD = DA = 7.
15. \overline{AC} bisects \angle A and \angle C, and nothing else is known.
16. AB = BC = CD, and nothing else is known.

Rectangle, rhombus or square

For each of 17 to 21, say whether ABCD must be a square, must be a rectangle, must be a rhombus, or need be none of these, naming the tests used.

17. ABCD is a parallelogram, AC = BD and \overline{AC} \perp \overline{BD}.
18. ABCD is a parallelogram, m\angle B = 90^\circ and AB = BC.
19. ABCD is a parallelogram, \overline{AC} \perp \overline{BD} and \angle A \cong \angle C.
20. ABCD is a parallelogram, AC = BD and AB = CD.
21. AC = BD and \overline{AC} \perp \overline{BD}, and nothing else is known.

Solving for unknowns

22. ABCD is a parallelogram with AC = 4x - 7 and BD = 2x + 11. Find the x that makes it a rectangle, and the length of each diagonal.
23. ABCD is a parallelogram with m\angle A = (5x - 20)^\circ and m\angle B = (3x + 40)^\circ. Find x, and decide whether ABCD is a rectangle.
24. ABCD is a parallelogram with m\angle A = (7y - 15)^\circ and m\angle B = (4y + 30)^\circ. Find y, and decide whether ABCD is a rectangle.
25. The diagonals of parallelogram ABCD meet at E, with CE = 2y + 3 and DE = y + 11. Find the y that makes ABCD a rectangle, and find AC and BD.
26. ABCD is a parallelogram with AB = 20, BC = 21 and AC = 29. Is it a rectangle? Find BD if you can.
27. PQRS is a parallelogram with PQ = 7, QR = 9 and PR = 12. Is it a rectangle?
28. ABCD is a parallelogram with AB = 4x - 3 and BC = 2x + 9. Find the x that makes it a rhombus, and find AB and the perimeter.
29. ABCD is a parallelogram and m\angle AED = (6y - 12)^\circ. Find the y that makes it a rhombus.
30. ABCD is a parallelogram with m\angle CAB = (2x + 11)^\circ and m\angle CAD = (4x - 17)^\circ. Find the x that makes it a rhombus, and then m\angle DAB and m\angle ABC.
31. ABCD is a parallelogram whose diagonals are perpendicular, with AE = 8 and BE = 15. Find AB, the perimeter, AC and BD.
32. ABCD is a rhombus and m\angle ABC = 124^\circ. Find m\angle BAD, m\angle BAC, m\angle ABD and m\angle AEB.

Coordinates

33. A(1, 1), B(7, 3), C(6, 6) and D(0, 4). Show that ABCD is a rectangle by the Equal Diagonals Test.
34. P(-3, 0), Q(1, -2), R(4, 2) and S(0, 4). Show that PQRS is a parallelogram, and decide whether it is a rectangle.
35. A(2, 1), B(6, 3) and C(4, 7) are three corners of parallelogram ABCD. Find D, and show that ABCD is a rectangle by the One Right Angle Test.
36. A(0, 0), B(5, -2), C(8, 0) and D(5, 6). Show that AC = BD, and that ABCD is not a rectangle.
37. A(1, 1), B(4, 5), C(9, 5) and D(6, 1). Use the Four Sides Test to show that ABCD is a rhombus, and show that it is not a square.
38. P(-1, 2), Q(3, 4), R(5, 0) and S(1, -2). Use the diagonals to show that PQRS is a rhombus. Is it also a square?
39. K(0, 0), L(4, 1), M(6, 5) and N(2, 4). Show that KLMN is a parallelogram but not a rhombus.
40. A(0, 0), B(13, 0) and D(5, k), with k > 0, are three vertices of parallelogram ABCD. Find the k that makes ABCD a rhombus, and the vertex C.
41. A(0, 0), B(3, 1), C(2, 4) and D(-1, 3). Using only the diagonals, show that ABCD is a square.

Proofs

Write each as a two-column proof or a paragraph.

42. Given. ABCD is a parallelogram whose diagonals meet at E, and \angle EAB \cong \angle EBA. Prove. ABCD is a rectangle.
43. Given. In quadrilateral ABCD, \angle A and \angle B are right angles and \overline{AD} \cong \overline{BC}. Prove. ABCD is a rectangle.
44. Given. ABCD is a parallelogram, M is the midpoint of \overline{AB}, and \overline{MD} \cong \overline{MC}. Prove. ABCD is a rectangle.
45. Given. The diagonals of quadrilateral ABCD are perpendicular bisectors of each other. Prove. ABCD is a rhombus.
46. Given. ABCD is a parallelogram and \overline{BD} bisects \angle ABC. Prove. \overline{AC} \perp \overline{BD}.
47. Given. In quadrilateral ABCD, \overline{AC} bisects \angle A and \angle C, and \overline{BD} bisects \angle B and \angle D. Prove. ABCD is a rhombus.
48. Given. ABCD is a rectangle, and P, Q, R and S are the midpoints of \overline{AB}, \overline{BC}, \overline{CD} and \overline{DA}. Prove. PQRS is a rhombus.

Reasoning

49. A clue says “neighboring angles of this parallelogram are supplementary”. Another says “neighboring angles of this parallelogram are congruent”. Explain why only the second one tells you the parallelogram is a rectangle.
50. Write the Equal Diagonals Test and its partner in the lesson as one “if and only if” statement. Explain why the word “parallelogram” in it cannot be replaced by “quadrilateral”.
51. A quadrilateral has congruent diagonals, and one of them cuts the other in half. Must it be a rectangle? Explain, with coordinates for an example if the answer is no.
52. Explain why a quadrilateral with three right angles is always a rectangle, while one with two right angles need not be, whichever two corners they are at.
53. The kite with perpendicular diagonals in the Boundaries has \overline{AC} bisecting \angle A. Try the proof of the Angle Bisector Test on it, with line 1 keeping only \angle BAC \cong \angle DAC. Which is the first line with no reason, and is the conclusion of line 5 true for the kite?
54. Explain why no quadrilateral that is not convex can have four congruent sides.
55. Write the Perpendicular Diagonals Test and the matching property of a rhombus together as one “if and only if” statement about parallelograms. Explain why the word “parallelogram” can be dropped from one direction but not from the other.
56. Can a diagonal of a parallelogram bisect the angle at one of its ends but not the angle at the other end? Explain.