Proving properties of shapes

AS · P1 · 15 min

"Show that $ABCD$ is a rectangle", "find the fourth vertex of the parallelogram", "find the area of the kite": these questions test whether you can turn the geometric properties of triangles and quadrilaterals into gradient, length and midpoint calculations. Nothing new is needed beyond lengths, gradients and midpoints and equations of lines. What earns the marks is choosing the right facts to check, setting them out clearly, and stating the conclusion.

The three tools

Every property of a polygon you will be asked about reduces to one of three calculations:

Geometric fact	Coordinate test
two sides are parallel	equal gradients
two sides are perpendicular	gradients multiply to $-1$ (or one horizontal, one vertical)
two sides are equal	equal lengths (compare the squared lengths)
two segments bisect each other	they have the same midpoint
a point lies on a line	its coordinates satisfy the equation

Comparing squared lengths is quicker and avoids surds: $AB^2 = 20$ and $BC^2 = 20$ shows $AB = BC$ directly.

What you need to show for each shape

A proof must show enough properties to force the shape, no more and no less. "Two sides are equal" does not prove a rhombus; "opposite sides are parallel" does not prove a rectangle.

Key result

Shape	Sufficient to show
Parallelogram	both pairs of opposite sides parallel, or the diagonals have the same midpoint
Rectangle	a parallelogram and one angle is $90^\circ$
Rhombus	a parallelogram and two adjacent sides equal, or the diagonals bisect each other at right angles
Square	a rectangle and two adjacent sides equal
Kite	two pairs of adjacent sides equal (one diagonal is then the perpendicular bisector of the other)
Trapezium	one pair of opposite sides parallel (and the other pair not parallel)
Isosceles triangle	two sides equal
Right-angled triangle	two sides perpendicular, or the side lengths satisfy Pythagoras

Properties of the diagonals

Diagonals give quick tests and quick constructions, so learn these:

In a parallelogram, the diagonals bisect each other (same midpoint).
In a rhombus, the diagonals bisect each other at right angles.
In a rectangle, the diagonals are equal and bisect each other.
In a square, the diagonals are equal and bisect each other at right angles.
In a kite, one diagonal is the perpendicular bisector of the other.

Setting out a proof

Sketch the points roughly, in order, so you can see which sides are adjacent and which are opposite.
Decide what is sufficient from the table.
Calculate each gradient, length or midpoint on its own line, with the substitution shown.
Write a sentence linking each result to a property: " $m_{PQ} = m_{SR}$ , so $PQ \parallel SR$ ".
Finish with a conclusion naming the shape.

Finding a missing vertex

The fourth vertex of a parallelogram

In parallelogram $ABCD$ , the step from $A$ to $B$ equals the step from $D$ to $C$ . So $D$ is found from $C$ by undoing the step from $A$ to $B$ . Equivalently, the diagonals share a midpoint, which gives

D = A + C - B \quad\text{(coordinate by coordinate)}

The order of the letters matters: in $ABCD$ , the vertices go round the shape, so $A$ and $C$ are opposite, and so are $B$ and $D$ .

Vertices from a diagonal

If you know two opposite vertices $A$ and $C$ of a rhombus, kite or square, the other diagonal lies along the perpendicular bisector of $AC$ . A further condition (the vertex lies on an axis, or on a given line, or the diagonals are equal) pins down the other vertices.

Areas

Choose the method by the shape:

Right angle present. Area of a right-angled triangle $= \tfrac{1}{2} \times$ (the two perpendicular sides). Area of a rectangle $=$ the product of adjacent sides.
Rhombus, kite or square. The diagonals are perpendicular, so area $= \tfrac{1}{2} \times d_1 \times d_2$ .
Isosceles triangle. The line from the apex to the midpoint of the base is perpendicular to the base, so it is the height.
Trapezium. Area $= \tfrac{1}{2}(a + b)h$ , where $h$ is the perpendicular distance between the parallel sides, found with a foot of a perpendicular.
Anything else. Enclose the shape in a rectangle with sides parallel to the axes, and subtract the right-angled triangles in the corners (the box method).

(1, 1) -- (6, 3) (6, 3) -- (2, 7) (2, 7) -- (1, 1) (1, 1) -- (6, 1) (6, 1) -- (6, 7) (6, 7) -- (1, 7) (1, 7) -- (1, 1)

The box method for the triangle $(1, 1)$ , $(6, 3)$ , $(2, 7)$ : the box is $5 \times 6 = 30$ , and the three corner triangles have areas $5$ , $8$ and $3$ , so the triangle has area $30 - 16 = 14$ .

Tip

For checking only: the area of a polygon with vertices $(x_1, y_1), \ldots, (x_n, y_n)$ taken in order is $\tfrac{1}{2}\left|\sum (x_i y_{i+1} - x_{i+1} y_i)\right|$ (the "shoelace" formula, with $(x_{n+1}, y_{n+1}) = (x_1, y_1)$ ). It is not on the syllabus, so do not rely on it alone in an answer, but it is a fast way to check your result.

Worked examples

A parallelogram from its diagonals

Show that $A(1, 2)$ , $B(6, 4)$ , $C(9, 9)$ and $D(4, 7)$ are the vertices of a parallelogram.

Solution

\text{Midpoint of } AC = \left(\frac{1 + 9}{2}, \frac{2 + 9}{2}\right) = (5,\ 5.5), \qquad \text{Midpoint of } BD = \left(\frac{6 + 4}{2}, \frac{4 + 7}{2}\right) = (5,\ 5.5)

The diagonals $AC$ and $BD$ have the same midpoint, so they bisect each other. Therefore $ABCD$ is a parallelogram.

A rectangle and its area

Show that $P(-2, 1)$ , $Q(2, 3)$ , $R(3, 1)$ and $S(-1, -1)$ form a rectangle, and find its area.

Solution

m_{PQ} = \frac{3 - 1}{2 + 2} = \frac{1}{2}, \quad m_{SR} = \frac{1 + 1}{3 + 1} = \frac{1}{2}, \quad m_{QR} = \frac{1 - 3}{3 - 2} = -2, \quad m_{PS} = \frac{-1 - 1}{-1 + 2} = -2

$PQ \parallel SR$ and $QR \parallel PS$ , so $PQRS$ is a parallelogram. Also $m_{PQ} \times m_{QR} = \tfrac{1}{2} \times (-2) = -1$ , so angle $PQR = 90^\circ$ . A parallelogram with a right angle is a rectangle.

PQ = \sqrt{4^2 + 2^2} = \sqrt{20}, \qquad QR = \sqrt{1^2 + 2^2} = \sqrt{5}

\text{Area} = \sqrt{20} \times \sqrt{5} = \sqrt{100} = 10

The fourth vertex

$A(-1, 2)$ , $B(3, 4)$ and $C(6, -1)$ are three vertices of the parallelogram $ABCD$ . Find the coordinates of $D$ .

Solution

The step from $B$ to $A$ is $(-4, -2)$ . Since $\overrightarrow{CD}$ is the same as $\overrightarrow{BA}$ in a parallelogram, $D = (6 - 4,\ -1 - 2) = (2, -3)$ .

Check with the diagonals: midpoint of $AC = (2.5, 0.5)$ and midpoint of $BD = \left(\tfrac{3 + 2}{2}, \tfrac{4 - 3}{2}\right) = (2.5, 0.5)$ . They agree.

A rhombus from one diagonal

The points $A(1, 2)$ and $C(7, 6)$ are opposite vertices of a rhombus $ABCD$ . The vertex $B$ lies on the $y$ -axis.

(a) Find the equation of the diagonal $BD$ .

(b) Find the coordinates of $B$ and $D$ .

Solution

(a) The diagonals of a rhombus bisect each other at right angles, so $BD$ is the perpendicular bisector of $AC$ . The midpoint of $AC$ is $M(4, 4)$ and $m_{AC} = \dfrac{4}{6} = \dfrac{2}{3}$ , so $BD$ has gradient $-\dfrac{3}{2}$ :

y - 4 = -\tfrac{3}{2}(x - 4) \quad\Rightarrow\quad 3x + 2y = 20

(b) $B$ is on the $y$ -axis, so $x = 0$ and $y = 10$ : $B = (0, 10)$ . $M$ is the midpoint of $BD$ , so the step $B \to M$ is $(4, -6)$ , and $D = (8, -2)$ .

(c)

AC = \sqrt{6^2 + 4^2} = \sqrt{52}, \qquad BD = \sqrt{8^2 + 12^2} = \sqrt{208}

\text{Area} = \tfrac{1}{2} \times \sqrt{52} \times \sqrt{208} = \tfrac{1}{2}\sqrt{10816} = \tfrac{1}{2}(104) = 52

Check: $AB^2 = 1 + 64 = 65$ and $BC^2 = 49 + 16 = 65$ , so the sides are equal, as they must be.

A trapezium and its area

The points $A(1, 1)$ , $B(9, 5)$ , $C(7, 9)$ and $D(3, 7)$ form a quadrilateral.

(a) Show that $ABCD$ is a trapezium.

(b) Find the coordinates of the foot of the perpendicular from $D$ to $AB$ .

Solution

(a)

m_{AB} = \frac{5 - 1}{9 - 1} = \frac{1}{2}, \quad m_{DC} = \frac{9 - 7}{7 - 3} = \frac{1}{2}, \quad m_{AD} = \frac{7 - 1}{3 - 1} = 3, \quad m_{BC} = \frac{9 - 5}{7 - 9} = -2

$AB \parallel DC$ , but $AD$ and $BC$ are not parallel. So $ABCD$ is a trapezium.

(b) Line $AB$ : $y - 1 = \tfrac{1}{2}(x - 1)$ , i.e. $x - 2y + 1 = 0$ . The perpendicular from $D$ has gradient $-2$ : $y - 7 = -2(x - 3)$ , i.e. $y = 13 - 2x$ . Substituting:

x - 2(13 - 2x) + 1 = 0 \quad\Rightarrow\quad 5x = 25 \quad\Rightarrow\quad x = 5,\ y = 3

The foot is $N(5, 3)$ .

(c) The height is $DN = \sqrt{2^2 + 4^2} = \sqrt{20} = 2\sqrt{5}$ . The parallel sides are $AB = \sqrt{80} = 4\sqrt{5}$ and $DC = \sqrt{20} = 2\sqrt{5}$ .

\text{Area} = \tfrac{1}{2}\left(4\sqrt{5} + 2\sqrt{5}\right) \times 2\sqrt{5} = \tfrac{1}{2} \times 6\sqrt{5} \times 2\sqrt{5} = 30

(1, 1) -- (9, 5) (9, 5) -- (7, 9) (7, 9) -- (3, 7) (3, 7) -- (1, 1) (3, 7) -- (5, 3)

A square from a diagonal

The points $A(1, 3)$ and $C(7, 5)$ are opposite vertices of a square $ABCD$ . Find the coordinates of $B$ and $D$ .

Solution

The diagonals of a square are equal, perpendicular and bisect each other. So $B$ and $D$ lie on the perpendicular bisector of $AC$ , at the same distance from the midpoint $M$ as $A$ is.

$M = (4, 4)$ and $m_{AC} = \tfrac{2}{6} = \tfrac{1}{3}$ , so the other diagonal has gradient $-3$ . Moving $1$ across and $-3$ up keeps you on it, so its points are $(4 + t,\ 4 - 3t)$ .

$MA^2 = 3^2 + 1^2 = 10$ . We need $MB^2 = 10$ as well:

t^2 + 9t^2 = 10 \quad\Rightarrow\quad t^2 = 1 \quad\Rightarrow\quad t = \pm 1

So the other two vertices are $(5, 1)$ and $(3, 7)$ . Going round in order $A, B, C, D$ (anticlockwise from $A$ ), $B = (5, 1)$ and $D = (3, 7)$ .

Check: $AB^2 = 16 + 4 = 20$ , $BC^2 = 4 + 16 = 20$ , and $m_{AB} = -\tfrac{1}{2}$ , $m_{BC} = 2$ , which are perpendicular.

Watch out

Proving too little. Four equal sides shows a rhombus, not a square. Equal diagonals in a parallelogram shows a rectangle, not a square. Check the table.

Wrong vertex order. In $ABCD$ , $A$ is joined to $B$ and $D$ , not to $C$ . Using $AC$ as a side instead of a diagonal is the most common error in fourth-vertex questions. Sketch first.

No conclusion. Calculations alone do not prove anything. End with a sentence: "so $ABCD$ is a rectangle".

Slanted heights. The height of a triangle or trapezium must be perpendicular to the base. A side that is not perpendicular cannot be used as the height.

Exam tip

"Show that" means every step. The result is given, so the marks are for the working. Show each gradient or length with the numbers substituted, then the comparison.
Use the squared length. Writing $AB^2 = 20 = BC^2$ is accepted and avoids surd slips.
Use the given order. If the question says "the quadrilateral $ABCD$ ", the sides are $AB$ , $BC$ , $CD$ and $DA$ .
Areas. Look first for a right angle or perpendicular diagonals; the box method is the fallback. State which lengths are perpendicular before multiplying.
Typical structure. These questions are often 6 to 9 marks over several parts: an equation of a line, an intersection, a missing vertex, then an area. Keep coordinates exact so later parts stay accurate.

Summary

Parallel: equal gradients. Perpendicular: product $-1$ . Equal: equal squared lengths. Bisect: same midpoint.
Parallelogram: opposite sides parallel or diagonals with a common midpoint. Add a right angle for a rectangle, equal adjacent sides for a rhombus, both for a square.
The fourth vertex of parallelogram $ABCD$ is $D = A + C - B$ .
Given two opposite vertices of a rhombus, kite or square, the other diagonal is the perpendicular bisector of the known one.
Areas: right angles, perpendicular diagonals ( $\tfrac{1}{2}d_1 d_2$ ), a perpendicular height, or the box method.
Always state the conclusion in words.

Practice questions

Question

Show that the points $(0, 0)$ , $(4, 2)$ , $(2, 6)$ and $(-2, 4)$ are the vertices of a square.
Show that the triangle with vertices $(-1, 3)$ , $(3, 5)$ and $(5, 1)$ is isosceles and right-angled, and find its area.
The points $A(2, 5)$ , $B(8, 7)$ , $C(10, 1)$ and $D(k, -1)$ are such that $AB$ is parallel to $DC$ . Find $k$ , and show that $ABCD$ is then a square.
$A(2, -1)$ , $B(5, 3)$ and $C(4, 7)$ are three vertices of a parallelogram $ABCD$ . Find $D$ .
Show that $P(0, 0)$ , $Q(4, 2)$ , $R(5, 5)$ and $S(2, 4)$ form a kite, and find its area.
Find the area of the triangle with vertices $(-2, 1)$ , $(4, 3)$ and $(1, 6)$ .
The points $A(-2, 1)$ and $C(4, 3)$ are opposite vertices of a rhombus $ABCD$ , and $B$ lies on the $y$ -axis. Find $B$ and $D$ , and show that $ABCD$ is in fact a square.
The points $A(-1, 3)$ , $B(7, 7)$ , $C(3, 10)$ and $D(-1, 8)$ form a quadrilateral. (a) Show that $AB$ is parallel to $DC$ . (b) Find the foot of the perpendicular from $D$ to $AB$ . (c) Find the area of $ABCD$ .
The points $A(0, 1)$ and $C(6, 5)$ are opposite vertices of a square $ABCD$ . Find the coordinates of the other two vertices and the area of the square.

Answers

Label them $O$ , $A$ , $B$ , $C$ . The steps round the shape are $(4, 2)$ , $(-2, 4)$ , $(-4, -2)$ , $(2, -4)$ , so each side has squared length $20$ : all four sides equal. $m_{OA} = \tfrac{1}{2}$ and $m_{AB} = \tfrac{4}{-2} = -2$ , product $-1$ , so there is a right angle. A rhombus with a right angle is a square.
The squared lengths are $4^2 + 2^2 = 20$ , $2^2 + 4^2 = 20$ and $6^2 + 2^2 = 40$ . Two sides are equal, so it is isosceles; $20 + 20 = 40$ , so by Pythagoras it is right-angled (at $(3, 5)$ ). Area $= \tfrac{1}{2} \times \sqrt{20} \times \sqrt{20} = 10$ .
$m_{AB} = \tfrac{2}{6} = \tfrac{1}{3}$ and $m_{DC} = \dfrac{1 + 1}{10 - k}$ . Setting $\dfrac{2}{10 - k} = \dfrac{1}{3}$ gives $k = 4$ . Then $m_{AD} = \dfrac{-1 - 5}{4 - 2} = -3$ and $m_{BC} = \dfrac{1 - 7}{10 - 8} = -3$ , so $AD \parallel BC$ and $ABCD$ is a parallelogram. $m_{AB} \times m_{AD} = -1$ , so it has a right angle. $AB^2 = 36 + 4 = 40$ and $AD^2 = 4 + 36 = 40$ , so adjacent sides are equal. Hence $ABCD$ is a square.
$D = A + C - B = (2 + 4 - 5,\ -1 + 7 - 3) = (1, 3)$ .
$PQ^2 = 16 + 4 = 20$ , $PS^2 = 4 + 16 = 20$ , $QR^2 = 1 + 9 = 10$ , $RS^2 = 9 + 1 = 10$ . Two pairs of adjacent sides are equal, so $PQRS$ is a kite. Diagonals: $PR = \sqrt{50}$ with gradient $1$ , $QS = \sqrt{8}$ with gradient $\tfrac{2}{-2} = -1$ , so they are perpendicular. Area $= \tfrac{1}{2}\sqrt{50}\sqrt{8} = \tfrac{1}{2}\sqrt{400} = 10$ .
Box from $x = -2$ to $4$ and $y = 1$ to $6$ : area $6 \times 5 = 30$ . Corner triangles: $\tfrac{1}{2}(6)(2) = 6$ , $\tfrac{1}{2}(3)(3) = 4.5$ , $\tfrac{1}{2}(3)(5) = 7.5$ , total $18$ . Area $= 30 - 18 = 12$ .
Midpoint of $AC$ is $(1, 2)$ and $m_{AC} = \tfrac{2}{6} = \tfrac{1}{3}$ , so $BD$ is $y - 2 = -3(x - 1)$ . At $x = 0$ , $y = 5$ : $B = (0, 5)$ . Then $D = (2, -1)$ . $AB^2 = 4 + 16 = 20$ and $BC^2 = 16 + 4 = 20$ ; $m_{AB} = 2$ and $m_{BC} = -\tfrac{1}{2}$ , product $-1$ . A rhombus with a right angle is a square. (Equivalently, $AC^2 = BD^2 = 40$ : equal diagonals.)
(a) $m_{AB} = \tfrac{4}{8} = \tfrac{1}{2}$ and $m_{DC} = \tfrac{2}{4} = \tfrac{1}{2}$ , so $AB \parallel DC$ . (b) $AB$ : $y - 3 = \tfrac{1}{2}(x + 1)$ , i.e. $x - 2y + 7 = 0$ . Perpendicular from $D$ : $y - 8 = -2(x + 1)$ , i.e. $y = 6 - 2x$ . Then $x - 2(6 - 2x) + 7 = 0$ , so $5x = 5$ , $x = 1$ , $y = 4$ . Foot $(1, 4)$ . (c) Height $\sqrt{2^2 + 4^2} = 2\sqrt{5}$ ; $AB = \sqrt{80} = 4\sqrt{5}$ ; $DC = \sqrt{20} = 2\sqrt{5}$ . Area $= \tfrac{1}{2}(6\sqrt{5})(2\sqrt{5}) = 30$ .
$M = (3, 3)$ and $m_{AC} = \tfrac{4}{6} = \tfrac{2}{3}$ , so the other diagonal has gradient $-\tfrac{3}{2}$ : its points are $(3 + 2t,\ 3 - 3t)$ . $MA^2 = 9 + 4 = 13$ , so $4t^2 + 9t^2 = 13$ , $t = \pm 1$ : the vertices are $(5, 0)$ and $(1, 6)$ . Area $= \tfrac{1}{2} \times AC^2 = \tfrac{1}{2} \times 52 = 26$ .

The three tools

What you need to show for each shape

Properties of the diagonals

Finding a missing vertex

The fourth vertex of a parallelogram

Vertices from a diagonal

Areas

Worked examples

Practice questions

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