If we want to make a quadrilateral from two congruent triangles we need to connect a pair of corresponding sides together. This makes sure that the points match up exactly.

Another way to think about this construction is to start with one triangle, and make a copy of the triangle with a rotation or a reflection. Try making some of your own quadrilaterals with the applet below:

Each mode always makes a kind of special quadrilateral.

If we use a reflection to make the second triangle, the quadrilateral we form always has two pairs of equal adjacent sides. This means a reflection always makes a kite.

If we use a rotation to make the second triangle, the quadrilateral we form always has two pairs of equal opposite sides. This means rotation always makes a parallelogram.

There are some special quadrilaterals that we can make by using either mode.

If we start with an isosceles triangle, then reflecting it across its base produces the same quadrilateral as rotating it around the midpoint of its base. The result is always a rhombus, which explains why rhombuses are both parallelograms and kites at the same time.

Can you make a rectangle? What about a square?

Practice questions

Question 1

Suppose we have $a$ $right-angled scalene triangle$ , then create another triangle by $rotating the original around the middle of its longest side$ .

If we join the triangles together, which of the following must be true of the resulting quadrilateral?

$It will always be a parallelogram but may not be a rectangle.$
A
$It will always be a rectangle but may not be a square.$
B
$It will always be a kite but may not be a parallelogram.$
C
$It will always be a square.$
D

Diagonals of special quadrilaterals

When we made our special quadrilaterals by joining two congruent triangles together, the sides of the triangles that were joined together formed one of the diagonals of the shape.

If we draw in the other diagonal of a quadrilateral we will make four triangles in total. Sometimes all four of these triangles will be congruent to each other, and sometimes there will only be two pairs of two. But in either case we can use these four triangles to investigate some interesting properties of the diagonals.

Let's have a look at the diagonals of a kite first. We could make the kite below by reflecting $\triangle DAB$△DAB across its side $DB$DB, and this side of the triangle becomes one of the diagonals of the kite we make. Drawing in the other diagonal $AC$AC splits the kite into four triangles, and we call the point where the diagonals cross $X$X:

Since $\triangle DAB\equiv\triangle DCB$△DAB≡△DCB, we know that $\angle DBA=\angle DBC$∠DBA=∠DBC. For the same reason we also know that $BA=BC$BA=BC.

Thinking about $\triangle ABX$△ABX and $\triangle CBX$△CBX, the side $BX$BX is common to both.

Putting this all together means we can conclude that $\triangle ABX\equiv\triangle CBX$△ABX≡△CBX by the SAS test.

Now that we know that these triangles are congruent, we can match up another pair of corresponding angles and conclude that $\angle BXA=\angle BXC$∠BXA=∠BXC.

Since these angles are supplementary, we know that $\angle BXA+\angle BXC=180^\circ$∠BXA+∠BXC=180°.

So the two angles $\angle BXA$∠BXA and $\angle BXC$∠BXC are equal, and they also add to $180^\circ$180°. This means they are both right angles. In other words,

The diagonals of a kite are perpendicular to each other.

Now let's investigate parallelograms. This time we rotate the triangle $\triangle DAB$△DAB around the midpoint $X$X of the side $DB$DB, so like before this side becomes a diagonal of the parallelogram we create. Drawing in the other diagonal $AC$AC splits the parallelogram into four triangles:

Since $\angle AXB$∠AXB and $\angle CXD$∠CXD are vertically opposite, they must be equal. The sides $AB$AB and $CD$CD are corresponding, so they must be equal too. These sides are also parallel, which means $\angle ABX$∠ABX and $\angle CDX$∠CDX are alternate angles on parallel lines.

Putting this all together tells us that $\triangle AXB\equiv\triangle CXD$△AXB≡△CXD by the AAS test.

Since $AX$AX and $CX$CX are corresponding sides in these congruent triangles, they must be equal. In other words, $X$X is not only the midpoint of the diagonal $BD$BD, it is also the midpoint of the diagonal $AC$AC. In other words,

The diagonals of a parallelogram bisect each other.

Finally, let's look at a rectangle. When we draw in the diagonals, we create many congruent triangles:

Let's look at the triangles $\triangle DBA$△DBA and $\triangle CAB$△CAB. We know that the side $AB$AB is common to both triangles, and we know $DA=CB$DA=CB as well. The angles $\angle DAB$∠DAB and $\angle CBA$∠CBA are both right angles, and are therefore equal.

We can therefore conclude that $\triangle DBA\equiv\triangle CAB$△DBA≡△CAB by the SAS test.

The sides $DB$DB and $CA$CA are corresponding sides, and must be equal. But these are also the diagonals of the rectangle. In other words,

The diagonals of a rectangle are equal in length.

There are other properties we could prove as well - at least one diagonal of a kite bisects the other, some of the quadrilaterals have diagonals that bisect the angles they pass through, and so on.

All of these properties can also be proved using the congruent triangles that form from drawing the diagonals.

Summary

These are the properties of the diagonals of special quadrilaterals:

	Kite	Parallelogram	Rectangle	Rhombus	Square
Diagonals are perpendicular	Yes			Yes	Yes
One diagonal bisects the other	Yes	Yes	Yes	Yes	Yes
Both diagonals bisect each other		Yes	Yes	Yes	Yes
One diagonal bisects the angles it passes through	Yes			Yes	Yes
Both diagonals bisect the angles they pass through				Yes	Yes
Diagonals are equal in length			Yes		Yes

Practice questions

Question 2

The triangles $\triangle ABC$△ABC and $\triangle CDA$△CDA are congruent.

Fill in the blanks, to state pairs of equal angles.

$\angle CAB=\angle\editable{}$∠CAB=∠

$\angle BCA=\angle\editable{}$∠BCA=∠
Which two of the following are true about the quadrilateral $ABCD$ABCD?
$AD\parallel BC$AD∥BC
Because $\angle ACD=\angle CAB$∠ACD=∠CAB
A
$AD\parallel BC$AD∥BC
Because $\angle DAC=\angle BCA$∠DAC=∠BCA
B
$AB\parallel DC$AB∥DC
Because $\angle ACD=\angle CAB$∠ACD=∠CAB
C
$AB\parallel DC$AB∥DC
Because $\angle DAC=\angle BCA$∠DAC=∠BCA
D
Which one of the following is true of the quadrilateral $ABCD$ABCD?
$ABCD$ABCD must be a $rhombus, but is not necessarily a square$ .
A
$ABCD$ABCD must be a $parallelogram, but is not necessarily a rhombus$ .
B
$ABCD$ABCD must be a $square$ .
C
$ABCD$ABCD must be a $rectangle, but is not necessarily a parallelogram$ .
D
Suppose that $\angle ACD=36^\circ$∠ACD=36° and $\angle BCA=44^\circ$∠BCA=44°. Find the size of $\angle ABC$∠ABC.

Question 3

The diagonals of a quadrilateral have these properties:

They $are$ equal in length
They $bisect$ each other
They $are not$ perpendicular

What is the quadrilateral's type?

$Kite$
A
$Parallelogram$
B
$Square$
C
$Rectangle$
D

Question 4

The diagonals of this $kite$ intersect at $T$T, splitting the $kite$ into four triangles.

The diagonal $PR$PR bisects the angles $\angle QRS$∠QRS and $\angle SPQ$∠SPQ.

Complete the following:

$\triangle RST$△RST and $\triangle\editable{}$△ are congruent triangles.
Which one of the following is true?
$\angle RTS=\angle TQR$∠RTS=∠TQR
A
$\angle RTS=\angle TRQ$∠RTS=∠TRQ
B
$\angle RTS=\angle RTQ$∠RTS=∠RTQ
C
Which one of the following is true?
$\angle RTS+\angle RTQ=180^\circ$∠RTS+∠RTQ=180°
A
$\angle RTS+\angle RTQ=90^\circ$∠RTS+∠RTQ=90°
B
$\angle RTS+\angle RTQ=360^\circ$∠RTS+∠RTQ=360°
C
Which two of the following are true about the diagonals of this kite?
$RP$RP bisects $SQ$SQ.
A
$RP$RP is perpendicular to $SQ$SQ.
B
$RS$RS is perpendicular to $RQ$RQ.
C
$SQ$SQ bisects $RP$RP.
D

7.05 Quadrilaterals

Constructing quadrilaterals from triangles

Practice questions

Question 1

Diagonals of special quadrilaterals

Practice questions

Question 2

Question 3

Question 4

What is Mathspace

About Mathspace