Properties of a parallelogram

Defining property. In parallelogram $ABCD$, side $AB$ is parallel to side $DC$, and side $AD$ is parallel to side $BC$ — written $AB \parallel DC$ and $AD \parallel BC$. Nothing else is assumed; every property below is a consequence of just this.

Four core properties (verified by measurement at Class 8 level):

• Opposite sides are equal: $AB = DC$ and $AD = BC$.
• Opposite angles are equal: $\angle A = \angle C$ and $\angle B = \angle D$.
• Adjacent angles are supplementary, since each pair sits between two parallel sides cut by a transversal.
• Diagonals bisect each other, but are not equal in length and do not meet at right angles in general.

For any adjacent pair such as $\angle A$ and $\angle B$, the supplementary relation is

where: $\angle A$ and $\angle B$ are the two angles of the parallelogram meeting at a shared side, and $180^\circ$ is a straight angle — the co-interior angle sum between the two parallel sides $AD$ and $BC$, cut by transversal $AB$. The same relation holds for every other adjacent pair around the shape.

Special parallelograms build on the same four properties, adding one extra fact about the diagonals:

• Rectangle = parallelogram with one angle $90^\circ$ — hence all four angles are $90^\circ$, since opposite angles are equal and adjacent angles are supplementary. Extra property: diagonals are equal ($AC = BD$), on top of bisecting each other.
• Rhombus = parallelogram with all four sides equal. Extra property: diagonals bisect each other at right angles, and each diagonal bisects the pair of vertex angles it passes through.
• Square = parallelogram that is both a rectangle and a rhombus. It carries every extra property at once: diagonals equal, bisecting, and perpendicular.

There is no formal proof expected at Class 8 — that comes with congruency in Class 9 — but the reasoning behind the four properties is easy to see, and seeing it makes the properties impossible to forget. Draw diagonal $AC$ in parallelogram $ABCD$. It splits the shape into triangles $ABC$ and $ACD$, and it also acts as a transversal crossing the two pairs of parallel sides. Because $AB \parallel DC$, the alternate angles across transversal $AC$ match: $\angle BAC = \angle DCA$. Because $AD \parallel BC$, the alternate angles across the same transversal also match: $\angle BCA = \angle DAC$. The two triangles share side $AC$, and two pairs of matching angles plus a shared side is enough to make the triangles identical in shape and size — so their remaining sides match too: $AB = DC$ and $AD = BC$. Opposite sides equal falls straight out of the parallel condition. The same pair of matching triangles explains equal opposite angles: angle $A$ of the parallelogram splits into the two angles that matched across triangles $ABC$ and $ACD$, and so does angle $C$, so $\angle A = \angle C$ (and by the same logic on the other diagonal, $\angle B = \angle D$). Adjacent angles being supplementary is just the co-interior angle fact for parallel lines cut by a transversal — $AD \parallel BC$ with transversal $AB$ gives $\angle A + \angle B = 180^\circ$ directly, no triangle needed. For the diagonals: once $AB = DC$ is established, drawing both diagonals creates two more matching triangles, $AOB$ and $COD$, again using alternate angles from the two pairs of parallel sides plus the now-equal side $AB = DC$. Matching triangles force $AO = OC$ and $BO = OD$ — the diagonals cut each other exactly in half. Nothing in this reasoning ever needed the diagonals to be equal in length or to cross at right angles, which is exactly why those two extra facts are not true for a general parallelogram — they only appear once you add the rectangle condition (equal angles) or the rhombus condition (equal sides) on top.

Problem. The diagonals of a rhombus $PQRS$ meet at $O$. If $PO = 6\text{ cm}$ and $QO = 8\text{ cm}$, find the side length $PQ$ and the perimeter of the rhombus.

Step 1. Identify which rhombus property applies. In a rhombus, the diagonals bisect each other at right angles, so at the point of intersection $O$: $\angle POQ = 90^\circ$, with $PO = 6\text{ cm}$ and $QO = 8\text{ cm}$ as the two segments meeting there.

Step 2. Recognise the right triangle. Triangle $POQ$ has a right angle at $O$, with legs $PO$ and $QO$, and its hypotenuse is $PQ$ — a full side of the rhombus, since $O$ lies on both diagonals and $PQ$ joins two adjacent vertices.

Step 3. Find the hypotenuse. The legs $6\text{ cm}$ and $8\text{ cm}$ form the familiar right-triangle triple $6$–$8$–$10$, so the hypotenuse is

Step 4. Use the rhombus side property to get the perimeter. All four sides of a rhombus are equal, so $PQ = QR = RS = SP = 10\text{ cm}$, and

Competitive stretch

If instead $\angle A = 3x + 10^\circ$ and $\angle B = 2x - 5^\circ$ were given for a parallelogram $ABCD$, the same co-interior fact used above, $\angle A + \angle B = 180^\circ$, gives $(3x+10^\circ)+(2x-5^\circ)=180^\circ$, so $5x = 175^\circ$ and $x = 35^\circ$, making $\angle A = 115^\circ$ and $\angle B = 65^\circ$. Opposite angles equal then hands over $\angle C = 115^\circ$ and $\angle D = 65^\circ$ with no further work, and the four angles sum to $360^\circ$ as a check.

• Assuming diagonals of a parallelogram are equal — only a rectangle (and square) has equal diagonals; a general parallelogram or a rhombus has diagonals that bisect each other but differ in length.
• Assuming diagonals of a parallelogram are perpendicular — only a rhombus (and square) crosses its diagonals at right angles; a general parallelogram or rectangle does not.
• Confusing "bisect each other" with "bisect the angles" — the diagonals of any parallelogram cut each other into equal halves, but they bisect the vertex angles only in a rhombus (or square).
• Swapping the equal-angle rule and the supplementary-angle rule — opposite angles are equal, adjacent angles are supplementary; mixing these up is the single most common slip when solving for unknown angles.

Ask the tutor: Ask your AI tutor for a construction-and-measurement check of these properties, or a mixed problem combining a rectangle's equal diagonals with a rhombus's perpendicular ones.