Proving Polygon Properties Analytically | IEB Grade 12 Maths

Analytical proofs ask you to show that a shape has a property using coordinates alone. The skill is choosing the shortest valid proof: which gradients or lengths to calculate and what conclusion they allow.

Proving the type of quadrilateral

Know the minimum condition for each shape:

Parallelogram: both pairs of opposite sides parallel, or diagonals bisect each other (same midpoint).
Rhombus: a parallelogram with two adjacent sides equal, or with perpendicular diagonals.
Rectangle: a parallelogram with one right angle, or with equal diagonals.
Square: a rhombus with one right angle, or a rectangle with two adjacent sides equal.
Kite: two pairs of adjacent sides equal.
Trapezium: one pair of opposite sides parallel.

Example. To find $D$ so that $A B C D$ is a parallelogram with $A (- 2; 1)$ , $B (3; 2)$ and $C (5; 6)$ , use equal midpoints of the diagonals:

$M_{A C} = (1, 5; 3, 5) = M_{B D} \Rightarrow D = (0; 5)$

Using gradients and distances in proofs

Gradients prove parallel or perpendicular sides. Distances prove equal sides. Midpoints prove bisection.

Example. $A (1; 1)$ , $B (4; 2)$ , $C (5; 5)$ and $D (2; 4)$ .

$m_{A B} = m_{D C} = \frac{1}{3}$ and $m_{A D} = m_{B C} = 3$ , so $A B C D$ is a parallelogram.
$A B = B C = 10$ , so it is a rhombus.
$m_{A B} \times m_{B C} = 1 \neq = - 1$ , so it has no right angle and is not a square.

Write a clear conclusion after each calculation, stating the property you used. A list of numbers with no conclusion does not earn the final mark.

Key terms

Parallelogram: A quadrilateral with both pairs of opposite sides parallel.
Rhombus: A parallelogram with all sides equal.
Rectangle: A parallelogram with a right angle.
Diagonal: A line joining opposite vertices of a polygon.

Common mistakes

Proving more than needed, or less: one pair of parallel sides does not prove a parallelogram.
Assuming a shape is a parallelogram before proving it.
Forgetting to write a conclusion after the calculations.
Using the midpoint method with the wrong pairing of vertices: diagonals join opposite vertices.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

4 marks total

routine

Prove that $A (1; 1)$ , $B (4; 2)$ , $C (5; 5)$ and $D (2; 4)$ are the vertices of a parallelogram.

Worked solution

Hidden - try the question first

Solution

$m_{A B} = \frac{2 - 1}{4 - 1} = \frac{1}{3}$ and $m_{D C} = \frac{5 - 4}{5 - 2} = \frac{1}{3}$ , so $A B ∥ D C$
$m_{A D} = \frac{4 - 1}{2 - 1} = 3$ and $m_{B C} = \frac{5 - 2}{5 - 4} = 3$ , so $A D ∥ B C$
Both pairs of opposite sides are parallel, so $A B C D$ is a parallelogram

Question 2

5 marks total

routine

$A (- 2; 1)$ , $B (3; 2)$ and $C (5; 6)$ are three vertices of parallelogram $A B C D$ . Determine the coordinates of $D$ .

Worked solution

Hidden - try the question first

Solution

Diagonals of a parallelogram bisect each other, so $M_{A C} = M_{B D}$
$M_{A C} = (\frac{- 2 + 5}{2}; \frac{1 + 6}{2}) = (\frac{3}{2}; \frac{7}{2})$
$\frac{3 + x}{2} = \frac{3}{2} \Rightarrow x = 0$
$\frac{2 + y}{2} = \frac{7}{2} \Rightarrow y = 5$
$D (0; 5)$

Question 3

7 marks total

complex

Use the parallelogram $A B C D$ from Question 1: $A (1; 1)$ , $B (4; 2)$ , $C (5; 5)$ , $D (2; 4)$ .

(a) Prove that $A B C D$ is a rhombus. (3)

(b) Show that the diagonals are perpendicular. (3)

Worked solution

Hidden - try the question first

Solution

(a) $M_{A C} = M_{B D}$
$B C = 1^{2} + 3^{2} = 10$
A parallelogram with two adjacent sides equal is a rhombus
(b) $m_{A C} = \frac{5 - 1}{5 - 1} = 1$
$m_{B D} = \frac{4 - 2}{2 - 4} = - 1$
$m_{A C} \times m_{B D} = - 1$ , so $A C ⊥ B D$
(c) No: $m_{A B} \times m_{B C} = \frac{1}{3} \times 3 = 1 \neq = - 1$ , so there is no right angle

Question 4

9 marks total

problem-solving

$P (- 3; 2)$ , $Q (5; 6)$ and $R (7; 2)$ are the vertices of $△ P QR$ .

(a) Prove that $\hat{P QR} = 90°$ . (3)

(b) Determine the coordinates of $S$ so that $P QR S$ is a rectangle. (2)

(d) Determine the centre of the circle through $P$ , $Q$ and $R$ . Give a reason. (2)

Worked solution

Hidden - try the question first

Solution

(a) $M_{A C} = M_{B D}$
$m_{QR} = \frac{2 - 6}{7 - 5} = - 2$
$m_{P Q} \times m_{QR} = - 1$ , so $P Q ⊥ QR$ and $m_{B D} = \frac{4 - 2}{2 - 4} = - 1$
(b) $M_{P R} = M_{QS}$ : $M_{P R} = (2; 2)$
$S = (2 \times 2 - 5; 2 \times 2 - 6) = (- 1; - 2)$
(c) $P Q = 80 = 45$ and $QR = 20 = 25$
Area $= 45 \times 25 = 40$ square units
(d) $P R$ is a diameter (chord subtends $90°$ )
Centre is $M_{P R} = (2; 2)$

FAQs

What is the quickest way to prove a parallelogram analytically?

Show the diagonals have the same midpoint. It needs only two midpoint calculations.

How do I prove a quadrilateral is a rectangle?

Prove it is a parallelogram, then show one angle is 90° using gradients with a product of $- 1$ , or show the diagonals are equal.

Is concurrency examined in IEB Grade 12 analytical geometry?

No. Concurrency of medians or altitudes is not examined.

Proving properties of polygons analytically

Proving the type of quadrilateral

Using gradients and distances in proofs

Key terms

Common mistakes

Practice questions

FAQs

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