Investigating Parallelograms Using Congruence
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- Investigating Parallelograms Using Congruence - Video - Parallelogram Proofs Watch
Theory
The properties of a parallelogram, such as equal opposite sides and diagonals that bisect each other, can all be proved with congruent triangles. This Year 10A maths page shows how to prove properties of parallelograms, rhombuses, rectangles and kites, how to prove that a quadrilateral is a parallelogram, and how to use the proved properties to find lengths and angles.
A parallelogram is a quadrilateral with both pairs of opposite sides parallel. A rhombus is a parallelogram with all four sides equal. A rectangle is a parallelogram with four right angles. A square is both a rhombus and a rectangle. A kite has two pairs of equal adjacent sides.
The other properties of these shapes, such as "the diagonals of a parallelogram bisect each other", are not part of the definition. They are proved by drawing a diagonal (or both diagonals) to make triangles, then proving two of the triangles congruent.
A congruence proof has three parts: name the two triangles; list three pairs of equal sides or angles, each with a reason; write the congruence statement with the test, e.g. \(\triangle ABC \equiv \triangle CDA\) (AAS). After that, "hence" means use the result just proved: matching parts of congruent triangles are equal.
Once a property has been proved, it can be used as a reason in later proofs, for example "opposite sides of a parallelogram are equal". The converse of a property swaps what is given and what is proved: "if the diagonals of a quadrilateral bisect each other, it is a parallelogram".
| Property | How it is proved |
|---|---|
| Parallelogram: opposite sides equal and opposite angles equal | one diagonal makes two triangles; alternate angles twice and the common diagonal give AAS |
| Parallelogram: diagonals bisect each other | the two triangles at the crossing point; alternate angles and opposite sides give AAS |
| Rectangle: diagonals are equal | common side, right angles and opposite sides give SAS |
| Rhombus: diagonals meet at right angles and bisect the vertex angles | equal sides, the common diagonal (or half-diagonal) and equal halves give SSS |
| Kite: the diagonal between the equal sides is perpendicular to the other and bisects it | SSS, then SAS (Example 1) |
How to prove a property of a quadrilateral
- Draw the diagonal (or diagonals) that splits the shape into the triangles you need. The question often names them.
- Mark the diagram: parallel sides give alternate angles; a diagonal is a common side; crossing diagonals give vertically opposite angles; equal sides come from the definition or from earlier properties.
- Write three pairs of equal parts, one per line, each with a reason. Keep the vertices in matching order.
- Conclude with \(\therefore\), the congruence statement and the test.
- Hence prove what was asked using matching sides or angles.
Finishing the proof
- Equal lengths or a midpoint: the two lengths are matching sides.
- Perpendicular: two matching angles on a straight line are each \(90^\circ\). Or, if matching angles show that two angles of a triangle add to \(90^\circ\), its third angle is \(90^\circ\).
- Parallel: a pair of matching angles are alternate angles, so the lines are parallel. Do this for both pairs of sides to show a parallelogram.
- Fill-the-gap proofs: each missing reason is a given fact, a standard angle fact, a property of the shape, the test, or "matching sides (angles) of congruent triangles".
Using the properties to find lengths and angles
- Name the property that links the parts, such as "diagonals bisect each other".
- Set the matching parts equal and solve the equation. If a quadratic gives two values, check each one in the other lengths and reject any that give a negative or mismatched length.
- Finish with Pythagoras' theorem, the angle sum of a triangle or a perimeter if needed.
(i) Prove that \(\triangle ABC \equiv \triangle ADC\).
(ii) Hence prove that the diagonals are perpendicular.
(i) In \(\triangle ABC\) and \(\triangle ADC\):
\(AB = AD\) (given)
\(CB = CD\) (given)
\(AC\) is common
\(\therefore\ \triangle ABC \equiv \triangle ADC\) (SSS)
(ii) \(\angle BAE = \angle DAE\) (matching angles of congruent triangles). Now use a second pair of triangles. In \(\triangle ABE\) and \(\triangle ADE\):
\(AB = AD\) (given)
\(\angle BAE = \angle DAE\) (just proved)
\(AE\) is common
\(\therefore\ \triangle ABE \equiv \triangle ADE\) (SAS), so \(\angle AEB = \angle AED\) (matching angles).
| \(\angle AEB + \angle AED\) | \(=\) | \(180^\circ\) (angles on a straight line) |
| \(\angle AEB\) | \(=\) | \(\angle AED = \mathbf{90^\circ}\) |
So \(AC \perp BD\). The same congruence gives \(BE = DE\), so the diagonal \(AC\) also bisects \(BD\).
Choose two triangles that each contain one diagonal and share a side. In \(\triangle ABC\) and \(\triangle BAD\):
\(AB\) is common
\(BC = AD\) (opposite sides of a parallelogram are equal)
\(AC = BD\) (given)
\(\therefore\ \triangle ABC \equiv \triangle BAD\) (SSS), so \(\angle ABC = \angle BAD\) (matching angles of congruent triangles).
These two angles are also co-interior angles, because \(AD \parallel BC\).
| \(\angle ABC + \angle BAD\) | \(=\) | \(180^\circ\) (co-interior angles, \(AD \parallel BC\)) |
| \(2\angle ABC\) | \(=\) | \(180^\circ\) |
| \(\angle ABC\) | \(=\) | \(\mathbf{90^\circ}\) |
A parallelogram with one right angle has four right angles (opposite angles are equal and neighbouring angles add to \(180^\circ\)), so \(\boldsymbol{ABCD}\) is a rectangle.
(i) \(ABCD\) is a parallelogram with diagonals meeting at \(O\). \(AO = (2x + 1)\) cm, \(OC = (5x - 11)\) cm, \(BO = (y + 3)\) cm and \(OD = (3y - 5)\) cm. Find the lengths of both diagonals.
(ii) Another parallelogram \(KLMN\) has \(KL = x^2\) cm, \(NM = (x + 12)\) cm and \(KN = (2x + 1)\) cm. Find \(x\) and the perimeter of \(KLMN\).
(i) The diagonals of a parallelogram bisect each other, so \(AO = OC\) and \(BO = OD\).
| \(2x + 1\) | \(=\) | \(5x - 11\) |
| \(x\) | \(=\) | \(4\) |
| \(y + 3\) | \(=\) | \(3y - 5\) |
| \(y\) | \(=\) | \(4\) |
\(AO = 9\) cm, so \(AC = \mathbf{18}\) cm. \(BO = 7\) cm, so \(BD = \mathbf{14}\) cm.
(ii) Opposite sides of a parallelogram are equal, so \(KL = NM\).
| \(x^2\) | \(=\) | \(x + 12\) |
| \(x^2 - x - 12\) | \(=\) | \(0\) |
| \((x - 4)(x + 3)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(4 \text{ or } x = -3\) |
If \(x = -3\), then \(KN = 2(-3) + 1 = -5\), which is impossible for a length, so reject it. With \(x = \mathbf{4}\): \(KL = 16\) cm and \(KN = 9\) cm.
| \(\text{perimeter}\) | \(=\) | \(2(16 + 9) = \mathbf{50}\text{ cm}\) |
The triangles \(ADE\) and \(CBF\) contain the two segments. Use the properties of a parallelogram as reasons. In \(\triangle ADE\) and \(\triangle CBF\):
\(AD = CB\) (opposite sides of a parallelogram are equal)
\(\angle DAE = \angle BCF\) (opposite angles of a parallelogram are equal)
\(AE = CF\) (given)
\(\therefore\ \triangle ADE \equiv \triangle CBF\) (SAS)
\(\therefore\ \boldsymbol{DE = BF}\) (matching sides of congruent triangles)
The angle at \(A\) is between \(AD\) and \(AE\), and the angle at \(C\) is between \(CB\) and \(CF\), so it is the included angle and SAS fits.
Common pitfalls
Frequently asked questions
How do you prove that a quadrilateral is a parallelogram?
Show that one of the tests holds: both pairs of opposite sides parallel or equal, both pairs of opposite angles equal, the diagonals bisect each other, or one pair of opposite sides equal and parallel. Most proofs draw a diagonal, prove two triangles congruent, then use equal alternate angles to show the sides are parallel.
Why do the diagonals of a parallelogram bisect each other?
The two triangles on opposite sides of the crossing point have equal alternate angles and equal opposite sides, so they are congruent (AAS). Their matching sides are the two halves of each diagonal.
What is the difference between a rhombus and a parallelogram?
A rhombus is a parallelogram with all four sides equal. It has every property of a parallelogram, and its diagonals also meet at right angles and bisect the vertex angles.
Are the diagonals of a rectangle equal?
Yes. The triangles made by each diagonal share a side, have equal right angles and equal opposite sides, so they are congruent (SAS) and the diagonals are matching sides.
Can a property of a shape be used as a reason?
Yes, once it has been proved, for example "opposite angles of a parallelogram are equal". Just do not use the property you are being asked to prove.