Proving Properties of Quadrilaterals
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Theory
To prove a property of a quadrilateral, split it with a diagonal, prove the two triangles congruent, then read the property off the matching parts. In parallelogram \(ABCD\), diagonal \(AC\) gives \(\triangle ADC \equiv \triangle CBA\), so \(\angle ADC = \angle CBA\). This Year 8 maths page shows how.
Start from the definitions. A parallelogram has both pairs of opposite sides parallel; a rhombus is a parallelogram with two adjacent sides equal; a rectangle is a parallelogram with a right angle; a square is a rectangle with two adjacent sides equal; a kite has two pairs of adjacent sides equal.
The usual three facts: a pair of equal alternate angles from parallel sides, a pair of equal sides, and the diagonal as a common side (SAS); or two pairs of alternate angles with the diagonal between them (AAS).
What the matching parts give: the opposite sides and opposite angles of a parallelogram are equal, and its diagonals bisect each other; the diagonals of a rhombus meet at right angles and bisect the vertex angles; the diagonals of a rectangle are equal; the axis of a kite bisects the other diagonal at right angles. A square has all of these.
Angles at the ends of one side of a parallelogram are not opposite angles: they are co-interior and add to \(180^\circ\).
Tests for a parallelogram. A quadrilateral is a parallelogram if both pairs of opposite sides are equal, or its diagonals bisect each other, or one pair of opposite sides is equal and parallel. Each is proved with congruent triangles and equal alternate angles.
Then use the properties: halve or double a bisected diagonal, use the \(360^\circ\) angle sum, and use Pythagoras’ theorem at the centre of a rhombus.
Properties proved with congruent triangles:
| Shape | Properties |
|---|---|
| parallelogram | opposite sides equal; opposite angles equal; diagonals bisect each other |
| rhombus | all of the above; diagonals meet at right angles and bisect the vertex angles |
| rectangle | parallelogram properties; diagonals equal |
| square | all the rhombus and rectangle properties |
| kite | one diagonal bisects the other at right angles |
Proving a property
- Draw the diagonal that makes two triangles holding the parts you need.
- List three facts: alternate angles (from parallel sides), given sides, the common diagonal.
- State the congruence and the test, with matching vertex order.
- Conclude the property from matching parts.
In \(\triangle AED\) and \(\triangle BFC\):
| Statement | Reason |
|---|---|
| \(AD = BC\) | given (hypotenuses) |
| \(AE = BF\) | given |
| \(\angle AED = \angle BFC\) | both \(90^\circ\) (given) |
| \(\triangle AED \equiv \triangle BFC\) | RHS |
| \(\angle ADE = \angle BCF\) | matching angles of congruent triangles |
In \(\triangle ADC\) and \(\triangle CBA\):
| Statement | Reason |
|---|---|
| \(AD = CB\) | opposite sides of a parallelogram |
| \(DC = BA\) | opposite sides of a parallelogram |
| \(AC = CA\) | common side |
| \(\triangle ADC \equiv \triangle CBA\) | SSS |
| \(\angle ADC = \angle CBA\) | matching angles of congruent triangles |
Note the order: \(A \leftrightarrow C\), \(D \leftrightarrow B\), \(C \leftrightarrow A\). So \(\angle ADC = \angle ABC\).
In \(\triangle KLN\) and \(\triangle MNL\):
| Statement | Reason |
|---|---|
| \(\angle KLN = \angle MNL\) | alternate angles, \(KL \parallel NM\) |
| \(\angle KNL = \angle MLN\) | alternate angles, \(KN \parallel LM\) |
| \(LN = NL\) | common side |
| \(\triangle KLN \equiv \triangle MNL\) | AAS |
| \(\angle LKN = \angle NML\) | matching angles of congruent triangles |
(i) \(PM\)
(ii) \(QS\)
(iii) \(\angle PSR\)
(iv) \(\angle QPS\).
| \(PM\) | \(=\) | \(46 \div 2 = 23\) |
| \(QS\) | \(=\) | \(2 \times 11 = 22\) |
| \(\angle QPS\) | \(=\) | \(180^\circ - 104^\circ = 76^\circ\) |
(i) \(23\) cm
(ii) \(22\) cm (the diagonals bisect each other)
(iii) \(104^\circ\) (opposite angles)
(iv) \(76^\circ\) (co-interior angles)
In \(\triangle ADC\) and \(\triangle BCD\):
| Statement | Reason |
|---|---|
| \(AD = BC\) | given |
| \(\angle ADC = \angle BCD\) | given |
| \(DC = CD\) | common side |
| \(\triangle ADC \equiv \triangle BCD\) | SAS |
| \(AC = BD\) | matching sides of congruent triangles |
(ii) A rhombus has diagonals \(16\) cm and \(30\) cm. Find its side length and perimeter.
| \(AE\) | \(=\) | \(12.6 \div 2 = 6.3\) |
| \(\text{side}^2\) | \(=\) | \(8^2 + 15^2 = 289\) |
| \(\text{side}\) | \(=\) | \(17~\text{cm}\) |
| \(\text{perimeter}\) | \(=\) | \(4 \times 17 = 68~\text{cm}\) |
(i) \(BD = 12.6\) cm (a rectangle’s diagonals are equal), \(AE = 6.3\) cm, \(\angle AEB = 90^\circ\) (a rhombus’s diagonals are perpendicular), \(\angle BAC = 45^\circ\) (the diagonal bisects the \(90^\circ\) angle).
(ii) The diagonals bisect each other at right angles, giving half-diagonals \(8\) cm and \(15\) cm: side \(17\) cm, perimeter \(68\) cm.
Common pitfalls
Frequently asked questions
How do you prove the opposite sides of a parallelogram are equal?
Draw a diagonal. Two pairs of alternate angles and the common diagonal give congruent triangles (AAS), so the matching sides are equal.
Why do the diagonals of a parallelogram bisect each other?
The diagonals make two congruent triangles (AAS, using alternate angles and equal opposite sides), so the matching halves are equal.
Do the diagonals of a rhombus meet at right angles?
Yes. Congruent triangles (SSS) show the two angles at the centre are equal, and together they make 180°, so each is 90°.
What makes a quadrilateral a parallelogram?
Both pairs of opposite sides parallel, or equal, or diagonals that bisect each other, or one pair of sides both equal and parallel.
Are the diagonals of a parallelogram equal?
Only when it is a rectangle.