Bisecting Diagonals, Equal Diagonals, Perpendicular Diagonals

Almost everything — if you know which of three questions to ask.

Given a quadrilateral whose diagonals bisect each other, you already know it is a parallelogram. Ask two more questions and the family narrows to one member:

- Are the diagonals equal? Then it is a rectangle.
- Do they meet at right angles? Then it is a rhombus.
- Both? Then it is a square.

Three yes-or-no questions, four answers. That is why diagonals are the thread running through this chapter: they identify a shape faster than measuring every side and angle does.

The trapezium and the kite sit outside this family — they are not parallelograms at all — and each has its own pair of properties worth knowing. This page covers the ICSE Class 8 Mathematics chapter on special types of quadrilaterals.

Opposite sides are equal and parallel, opposite angles are equal, adjacent angles are supplementary, and the diagonals bisect each other.

Set out carefully, for a parallelogram $ABCD$:

- $AB=DC$ and $AD=BC$, with $AB\parallel DC$ and $AD\parallel BC$
- $\angle A=\angle C$ and $\angle B=\angle D$
- $\angle A+\angle B=180^{\circ }$, and likewise for every adjacent pair
- The diagonals cut each other exactly in half — but they are not equal
- Each diagonal divides the parallelogram into two congruent triangles

Worked example 1 — one angle gives all four. *In parallelogram $ABCD$, $\angle A=70^{\circ }$.*

Since opposite angles are equal, $\angle C=70^{\circ }$. Adjacent angles are supplementary, so

$$\angle B=180-70=110^{\circ },\angle D=110^{\circ }$$

Check: $70+110+70+110=360^{\circ }$. Correct.

Worked example 2 — angles in a ratio. *Two adjacent angles of a parallelogram are in the ratio $2:3$.*

Adjacent angles sum to $180^{\circ }$, so five parts share $180^{\circ }$ and one part is $36^{\circ }$:

$$\text{angles}=72^{\circ }\text{ and }108^{\circ }$$

The four angles are $72^{\circ }$, $108^{\circ }$, $72^{\circ }$, $108^{\circ }$, totalling $360^{\circ }$. Correct.

Worked example 3 — a perimeter problem. *A parallelogram has perimeter $44$ cm, and one side is $3$ cm longer than the other.*

Only two different side lengths exist, since opposite sides are equal. With the shorter side $x$ cm, the longer is $x+3$ cm and the perimeter is

$$2(x+(x+3))=44\Rightarrow 2(2x+3)=44\Rightarrow 4x+6=44$$

$$4x=38\Rightarrow x=9.5$$

The sides are $9.5$ cm and $12.5$ cm. Check: $2(9.5+12.5)=2\times 22=44$ cm, and the difference is $3$ cm. Correct.

Expand that bracket carefully. $2(2x+3)$ is $4x+6$, not $4x+10$ — and the perimeter check catches that slip in one line, which is the argument for always doing it.

One right angle is enough. If a parallelogram has a single angle of $90^{\circ }$, its adjacent angle is $180-90=90^{\circ }$ too, and so is every other. A parallelogram with one right angle is a rectangle — you never need to be told about all four.

A rectangle has equal diagonals; a rhombus has perpendicular ones; a square has both.

All three are parallelograms, so all three inherit bisecting diagonals. What separates them is what happens on top of that:

- Rectangle: all angles $90^{\circ }$; diagonals equal and bisecting, but not perpendicular
- Rhombus: all sides equal; diagonals perpendicular and bisecting, but not equal
- Square: all sides equal and all angles $90^{\circ }$; diagonals equal, perpendicular, bisecting, and they also bisect the corner angles

Worked example 1 — a rhombus from its diagonals. *A rhombus has diagonals of $16$ cm and $12$ cm. Find its side and area.*

The diagonals bisect each other at right angles, so they create four right-angled triangles with legs of $8$ cm and $6$ cm. By Pythagoras the side is

$$\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10\text{ cm}$$

The perimeter is $40$ cm, and the area of a rhombus is half the product of its diagonals:

$$\text{area}=\frac{1}{2}\times 16\times 12=96{\text{ cm}}^2$$

Worked example 2 — working the other way. *A rhombus has side $13$ cm and one diagonal $24$ cm. Find the other diagonal.*

Half the known diagonal is $12$ cm, and the side is the hypotenuse:

$$\text{half of other diagonal}=\sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5\text{ cm}$$

So the other diagonal is $10$ cm, and the area is $\frac{1}{2}\times 24\times 10=120{\text{ cm}}^2$.

Worked example 3 — a rectangle's diagonal. *A rectangle measures $8$ cm by $6$ cm.*

$$\text{diagonal}=\sqrt{8^2+6^2}=10\text{ cm}$$

Both diagonals are $10$ cm, since a rectangle's diagonals are equal. They are not perpendicular — they meet at the centre at an angle determined by the shape, and only when the rectangle is a square does that angle become $90^{\circ }$.

Worked example 4 — a square's diagonal. *A square has side $5$ cm.*

$$\text{diagonal}=\sqrt{5^2+5^2}=\sqrt{50}=5\sqrt{2}\approx 7.07\text{ cm}$$

The misconception this topic exists to correct. Diagonals bisect each other at right angles does not mean square — it means rhombus. A square needs the diagonals to be equal as well. Similarly, diagonals are equal gives a rectangle, not a square. Students who learn the square's properties first tend to attribute them all to the rhombus and the rectangle, and questions are written to catch precisely that.

A trapezium has one pair of parallel sides; a kite has two pairs of adjacent equal sides.

Neither is a parallelogram, so neither has bisecting diagonals — and that is the first thing to hold on to.

Trapezium. Exactly one pair of sides is parallel. Because a pair of parallel lines is crossed by each of the non-parallel sides, the two angles along each slanted side are supplementary.

Worked example 1. *In trapezium $ABCD$ with $AB\parallel DC$, $\angle A=75^{\circ }$. Find $\angle D$.*

$\angle A$ and $\angle D$ lie on the same slanted side $AD$ between the parallels, so

$$\angle D=180-75=105^{\circ }$$

Isosceles trapezium. The non-parallel sides are equal. This adds three properties: the base angles are equal (both angles on the same parallel side), and the diagonals are equal.

Worked example 2. *In an isosceles trapezium $ABCD$ with $AB\parallel DC$ and $\angle A=60^{\circ }$, find all four angles.*

Base angles on $AB$ are equal, so $\angle B=60^{\circ }$. Then

$$\angle D=180-60=120^{\circ },\angle C=120^{\circ }$$

Check: $60+60+120+120=360^{\circ }$. Correct.

Kite. Two pairs of adjacent sides are equal — not opposite sides, which is what makes it different from a parallelogram. Its properties:

- One pair of opposite angles is equal (the pair where the unequal sides meet)
- One diagonal is the perpendicular bisector of the other
- That same diagonal is a line of symmetry, and it bisects the two angles it passes through

Worked example 3. *Kite $ABCD$ has $AB=AD=6$ cm and $CB=CD=10$ cm, with $\angle A=110^{\circ }$ and $\angle B=\angle D=85^{\circ }$. Find $\angle C$ and the perimeter.*

$$\angle C=360-110-85-85=80^{\circ }$$

$$\text{perimeter}=6+6+10+10=32\text{ cm}$$

Check: $110+85+80+85=360^{\circ }$. Correct.

**The equal angles are $\angle B$ and $\angle D$, not $\angle A$ and $\angle C$. In a kite the equal pair is the one between** the unequal sides — here $B$ and $D$, where a $6$ cm side meets a $10$ cm side. The angles at $A$ and $C$ sit between two equal sides and are generally different from each other. Getting that pair the wrong way round is the standard kite error.

How a rhombus fits in. A rhombus satisfies the kite definition — all four sides equal means two pairs of adjacent equal sides — so every rhombus is a kite, though most kites are not rhombuses. That also explains why the rhombus has perpendicular diagonals: it inherits the property from being a kite.

Quote the property you are using at every step, and work from the definition rather than from the picture.

Worked example 1 — a short proof. Show that if the diagonals of a parallelogram are equal, it is a rectangle.

In parallelogram $ABCD$, the diagonals bisect each other at $O$, so $AO=OC$ and $BO=OD$. If the diagonals are equal then $AC=BD$, and halving gives $AO=BO=CO=DO$ — all four half-diagonals equal.

So $O$ is the same distance from all four vertices. Triangles $AOB$ and $BOC$ are then isosceles, and working through the base angles forces each corner angle of the quadrilateral to be $90^{\circ }$. A parallelogram with right angles is a rectangle.

Worked example 2 — a numerical deduction. *The diagonals of a parallelogram are $10$ cm and $14$ cm. What are the lengths from the centre to each vertex?*

Since the diagonals bisect each other, the halves are $5$ cm and $7$ cm. So the centre is $5$ cm from two opposite vertices and $7$ cm from the other two. They are not all equal — if they were, the diagonals would be equal and the figure would be a rectangle.

Worked example 3 — using the angle sum with algebra. *The angles of a quadrilateral are $x$, $2x-10^{\circ }$, $x+30^{\circ }$ and $2x-20^{\circ }$. Find $x$, and decide whether the figure could be a parallelogram.*

$$x+(2x-10)+(x+30)+(2x-20)=360$$

$$6x+0=360\Rightarrow x=60$$

The four angles, in order, are $60^{\circ }$, $110^{\circ }$, $90^{\circ }$ and $100^{\circ }$. Verify the total: $60+110+90+100=360^{\circ }$. Correct.

Now test the parallelogram condition. Opposite angles would have to be equal, but the first and third are $60^{\circ }$ and $90^{\circ }$, and the second and fourth are $110^{\circ }$ and $100^{\circ }$. Neither pair matches, so this is not a parallelogram — even though one of its angles happens to be a right angle.

The habit that earns proof marks. Write the reason beside each statement — diagonals of a parallelogram bisect each other, adjacent angles of a parallelogram are supplementary, *angles of a quadrilateral sum to $360^{\circ }$*. A chain of true statements with no reasons attached reads as a guess, and a reason you cannot name is a property you have not actually checked.