12.10 Quadrilaterals built from triangles

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.

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 transformation.

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.

Examples

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 this kite by reflecting \triangle DAB across its side DB, and this side of the triangle becomes one of the diagonals of the kite we make. Drawing in the other diagonal AC splits the kite into four triangles, and we call the point where the diagonals cross X.

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

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

Putting this all together means we can conclude that \triangle ABX \equiv \triangle 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.

Since these angles are supplementary, we know that \angle BXA + \angle BXC = 180\degree.

So the two angles \angle BXA and \angle BXC are equal, and they also add to 180\degree. 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 around the midpoint X of the side DB, so like before this side becomes a diagonal of the parallelogram we create. Drawing in the other diagonal AC splits the parallelogram into four triangles.

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

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

Since AX and CX are corresponding sides in these congruent triangles, they must be equal. In other words, X is not only the midpoint of the diagonal BD, it is also the midpoint of the diagonal 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 and \triangle CAB. We know that the side \triangle DBA is common to both triangles, and we know DA = CB as well. The angles \angle DAB and \angle CBA are both right angles, and are therefore equal.

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

The sides DB and 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.

Examples

Example 2

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

a

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

\angle CAB = \angle ⬚ \\ \angle BCA = \angle ⬚

Worked Solution

Create a strategy

Remember that corresponding angles in congruent triangles are equal.

Apply the idea

The corresponding angle of \angle CAB in \triangle CDA is \angle ACD.

The corresponding angle of \angle BCA in \triangle CDA is \angle DAC.

So we have the pairs of equal angles:

\angle CAB = \angle ACD \\ \angle BCA = \angle DAC

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b

Which two of the following are true about the quadrilateral ABCD?

A

AD \parallel BC because \angle ACD = \angle CAB

B

AD \parallel BC because \angle DAC = \angle BCA

C

AB \parallel DC because \angle ACD = \angle CAB

D

AB \parallel DC because \angle DAC = \angle BCA

Worked Solution

Create a strategy

Consider AC as a transversal and identify the alternate angles.

Apply the idea

Let's look at AC as a transversal between AD and BC:

The alternate angles on this transversal are \angle DAC and \angle BCA which we know are equal from part (a).

So: AD \parallel BC because \angle DAC = \angle BCA.

Now looking at AC as a transversal between AB and DC, then the alternate angles on this transversal are \angle ACD and \angle CAB which we know are equal from part (a).

So: AB \parallel DC because \angle ACD = \angle CAB.

Options B and C are the correct answers.

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c

Which one of the following is true of the quadrilateral ABCD?

A

ABCD must be a rhombus, but is not necessarily a square.

B

ABCD must be a parallelogram, but is not necessarily a rhombus.

C

ABCD must be a square.

D

ABCD must be a rectangle, but is not necessarily a parallelogram.

Worked Solution

Create a strategy

Use the diagram below as guide to help us find the name of the quadrilateral with features matching ABCD.

Apply the idea

We know from the parts (a) and (b) that the quadrilateral ABCD has two pairs of opposite parallel sides and the opposite sides are equal.

The type of quadrilateral that is defined by this features is a parallelogram.

So ABCD must be a parallelogram, but is not necessarily a rhombus.

So option B is the correct answer.

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d

Suppose that \angle ACD = 36\degree and \angle BCA = 44\degree. Find the size of \angle ABC.

Worked Solution

Create a strategy

Remember that co-interior angles on parallel lines add to 180\degree.

Apply the idea

Since AB and DC are parallel, the angles \angle ACD,\, \angle BCA and \angle ABC are co-interior angles. So we have:

\displaystyle \angle ACD + \angle BCA + \angle ABC	\displaystyle =	\displaystyle 180	Co-interior angles on parallel lines
\displaystyle 36 + 44 + \angle ABC	\displaystyle =	\displaystyle 180	Substitute the values
\displaystyle 80 + \angle ABC	\displaystyle =	\displaystyle 180	Add the like terms
\displaystyle \angle ABC	\displaystyle =	\displaystyle 180-80	Subtract 80 from both sides
\displaystyle \angle ABC	\displaystyle =	\displaystyle 100\degree	Evaluate

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Example 3

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

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

a

Which triangle is congruent to \triangle RST?

Worked Solution

Create a strategy

Look for the triangle that has the same markings and common sides.

Apply the idea

Based on the markings given in the diagram we have RS=RQ and \angle SRT = \angle QRT. The two triangles also share a common side, RT.

So \triangle RST \equiv \triangle RQT by SAS.

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b

Which one of the following is true?

A

\angle RTS =\angle TQR

B

\angle RTS=\angle TRQ

C

\angle RTS=\angle RTQ

Worked Solution

Create a strategy

Find the angle in \triangle RQT that corresponds to the angle \angle RTS in \triangle RST.

Apply the idea

The angle in \triangle RQT that corresponds to the angle \angle RTS in \triangle RST is \angle RTQ.

So \angle RTS = \angle RTQ.

Option C is the correct answer.

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c

Which one of the following is true?

A

\angle RTS+\angle RTQ = 180\degree

B

\angle RTS+\angle RTQ = 90\degree

C

\angle RTS+\angle RTQ =360\degree

Worked Solution

Create a strategy

Identify the kind of angle they form together.

Apply the idea

Looking at the two angles, \angle RTS and \angle RTQ, in the diagram we can see that together they form a straight angle.

A straight angle measures 180\degree so we have: \angle RTS+\angle RTQ = 180\degree.

So option A is the correct answer.

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d

Which two of the following are true about the diagonals of this kite?

A

RP bisects SQ.

B

RP is perpendicular to SQ.

C

RS is perpendicular to RQ.

D

SQ bisects RP.

Worked Solution

Create a strategy

Use the findings from the previous parts.

Apply the idea

We know from parts (b) and (c) that \angle RTS and \angle RTQ are equal, and add to 180\degree. So we have:

\displaystyle \angle RTS+\angle RTQ	\displaystyle =	\displaystyle 180
\displaystyle \angle RTS+\angle RTS	\displaystyle =	\displaystyle 180	Replace \angle RTQ with \angle RTS
\displaystyle 2 (\angle RTS)	\displaystyle =	\displaystyle 180	Simplify
\displaystyle \dfrac{2 (\angle RTS)}{2}	\displaystyle =	\displaystyle \dfrac{180}{2}	Divide both sides by 2
\displaystyle \angle RTS	\displaystyle =	\displaystyle 90\degree	Evaluate

So we can say that RP is perpendicular to SQ.

We know in part (a) that \triangle RST \equiv \triangle RQT so ST and QT are corresponding sides. So ST = QT. We can say that RP bisects SQ.

So options A and B are the correct answers.

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