Properties of parallelograms

Take any quadrilateral, force both pairs of opposite sides to run parallel, and something surprising happens: sides, angles, and diagonals all snap into fixed relationships automatically. That quadrilateral is a parallelogram, and Class 9 geometry spends a whole chapter mining the consequences of that one parallel condition. Opposite sides turn out equal, opposite angles turn out equal, and the two diagonals always cut each other exactly in half — none of this is assumed, all of it is proved from a single diagonal and the ASA congruence rule. Better still, every one of these facts runs in reverse: if you can show a quadrilateral has equal opposite sides, or equal opposite angles, or bisecting diagonals, you may conclude it is a parallelogram without checking parallelism directly. This gives you a toolkit of forward facts and converse tests that recurs constantly — in proofs, in numerical problems, and in the very next chapter on areas of parallelograms and triangles.

The core insight

In a parallelogram, one diagonal always splits it into two congruent triangles by ASA, and from that single congruence follow all the standard properties — equal opposite sides, equal opposite angles, and diagonals that bisect each other — each of which also holds as a usable converse test.

The core idea

A parallelogram $ABCD$ is a quadrilateral with $AB \parallel DC$ and $AD \parallel BC$ . Everything else is a consequence of this parallel condition, proved using alternate-angle facts (from parallel lines cut by a transversal) together with triangle congruence — mainly the ASA criterion. The four standard properties, all provable and usable directly:

Opposite sides are equal: $AB = CD$ and $AD = BC$ .
Opposite angles are equal: $\angle A = \angle C$ and $\angle B = \angle D$ .
Consecutive angles are supplementary: since $AD$ is a transversal cutting the parallel lines $AB$ and $DC$ , co-interior angles give $\angle A + \angle D = 180°$ , and likewise for every adjacent pair.
Diagonals bisect each other: if diagonals $AC$ and $BD$ meet at $O$ , then $AO = OC$ and $BO = OD$ .

Each property also has a true converse, giving four independent tests for "is this quadrilateral a parallelogram":

\text{both pairs of opposite sides equal} \iff \text{parallelogram (with the parallel definition)}.

where: the double arrow records that the property and its converse are both valid — you may use whichever direction a problem needs. The other three converses read the same way: equal opposite angles; one pair of opposite sides both equal and parallel; or diagonals bisecting each other — any single one of these four conditions, once verified, is by itself enough to conclude the quadrilateral is a parallelogram.

Figure

Parallelogram ABCD with diagonal AC drawn showing the two congruent triangles and diagonals meeting at O

The core proof

Derivation ADiagonal AC divides a parallelogram into two congruent triangles

The goal is to show that in parallelogram $ABCD$ with $AB \parallel DC$ and $AD \parallel BC$ , the diagonal $AC$ produces $\triangle ABC \cong \triangle CDA$ , using only the parallel-line angle facts and the ASA congruence criterion — this single result is the seed from which equal opposite sides, equal opposite angles, and (with a short extra argument) the bisecting-diagonals property are all derived in the standard course sequence.

Step 1. Treat $AC$ as a transversal cutting the parallel lines $AB$ and $DC$ . Alternate interior angles are equal, so $\angle BAC = \angle DCA$ .

Step 2. Treat $AC$ as a transversal cutting the other pair of parallel lines, $AD$ and $BC$ . Alternate interior angles again give $\angle DAC = \angle BCA$ .

Step 3. Note the side $AC$ is shared between the two triangles: $AC = CA$ (common side), and it is the side included between the two angle pairs identified in Steps 1 and 2 for each triangle.

Step 4. Match up the pieces for $\triangle ABC$ and $\triangle CDA$ : angle $\angle BAC = \angle DCA$ , the included side $AC = CA$ , and angle $\angle BCA = \angle DAC$ . This is exactly angle-side-angle, so by the ASA criterion,

\triangle ABC \cong \triangle CDA.

where: the vertex order on both sides of the congruence statement tells you the correspondence — $A \leftrightarrow C$ , $B \leftrightarrow D$ , $C \leftrightarrow A$ — which is what lets you read off matching sides and angles correctly in the next step.

Step 5. Read off corresponding parts (CPCT): $AB = CD$ , $BC = DA$ — opposite sides equal — and $\angle B = \angle D$ . Combining $\angle B = \angle D$ with the equal alternate angles from Steps 1–2 (which split $\angle A$ and $\angle C$ into matching pieces) gives $\angle A = \angle C$ too, so opposite angles are equal as well. The bisecting-diagonals property follows separately, by applying this same congruent-triangles idea to the triangles formed at the intersection point $O$ of both diagonals.

Worked example

Problem. In parallelogram $PQRS$ , $\angle P = 3x + 10°$ and $\angle Q = 2x - 5°$ . Find all four angles, and also state which sides of $PQRS$ must be equal.

Step 1. Identify the angle relationship. $PQ \parallel SR$ and $PS$ is a transversal, so $\angle P$ and $\angle Q$ are consecutive (co-interior) angles of the parallelogram, which are supplementary:

\angle P + \angle Q = 180°.

Step 2. Substitute the given expressions and solve for $x$ :

(3x + 10°) + (2x - 5°) = 180° \implies 5x + 5° = 180° \implies 5x = 175° \implies x = 35°.

Step 3. Compute the two given angles: $\angle P = 3(35°) + 10° = 115°$ and $\angle Q = 2(35°) - 5° = 65°$ .

Step 4. Use the opposite-angles-equal property to get the remaining two angles: $\angle R = \angle P = 115°$ and $\angle S = \angle Q = 65°$ . Check: $115° + 65° + 115° + 65° = 360°$ , matching the angle sum of any quadrilateral, so the answer is consistent.

Step 5. By the opposite-sides-equal property (proved above via ASA), $PQ = SR$ and $PS = QR$ — this holds regardless of the numeric angle values, since it follows purely from $PQRS$ being a parallelogram.

Final answer: $\angle P = 115°$ , $\angle Q = 65°$ , $\angle R = 115°$ , $\angle S = 65°$ , with $PQ = SR$ and $PS = QR$ .

Competitive stretch

Extending Worked Example 2 from the base article: if diagonals of quadrilateral $ABCD$ meet at $O$ with $AO = OC = 5\ \text{cm}$ and $BO = OD = 7\ \text{cm}$ , the converse "diagonals bisect each other $\implies$ parallelogram" immediately gives $ABCD$ a parallelogram, with $AC = 10\ \text{cm}$ and $BD = 14\ \text{cm}$ . Since $AC \ne BD$ , this parallelogram is not a rectangle — a useful reminder that "diagonals bisect each other" is a weaker condition than "diagonals are equal," and the two should never be interchanged.

Common traps

Assuming diagonals are equal. Diagonals bisecting each other means each is cut into two equal halves by the other — it says nothing about the two diagonals having the same total length; that stronger condition only holds for rectangles.
Mixing up opposite and consecutive angles. Opposite angles are equal; consecutive (adjacent) angles are supplementary. Swapping the two relations is the most common algebra slip when setting up equations like in the worked example above.
Using a converse without checking the exact condition. "One pair of opposite sides equal and parallel" is a valid test, but "one pair of opposite sides merely equal" (without also parallel) is not enough — that quadrilateral could instead be an isosceles trapezium.
Forgetting ASA needs the included angle. In the diagonal-congruence proof, the two equal angles must sit adjacent to the shared side $AC$ ; mislabeling which angle is "included" invalidates the ASA match.

Ask the tutor: Want to walk through the diagonals-bisect-each-other proof step by step, or try another angle-finding problem like the worked example above?

Why does diagonals bisecting each other prove a quadrilateral is a parallelogram, and not just any four-sided shape

Because that specific condition forces both pairs of opposite sides to be parallel, which is the actual definition of a parallelogram. If $O$ is the midpoint of both diagonals, the two triangles formed at $O$ (say $\triangle AOB$ and $\triangle COD$ ) are congruent by SAS, using the equal diagonal halves and equal vertical angles at $O$ . That congruence gives $AB = CD$ and, crucially, $\angle OAB = \angle OCD$ , which are alternate angles that force $AB \parallel CD$ . Repeating with the other pair of triangles gives $AD \parallel BC$ . Both pairs parallel is exactly the definition, so the quadrilateral must be a parallelogram — no extra shape can satisfy this bisecting condition without also being one.

Do I need to memorize four separate proofs for the four properties

No — in the NCERT course, only the diagonal-congruence proof (opposite sides and angles equal) is typically asked to be proved in full; the diagonals-bisect-each-other property and all four converses are usually stated and used directly, without re-proving them from scratch each time. Understanding the one ASA argument in the derivation above, and knowing which property or converse to reach for in a given problem, covers almost everything Class 9 exams ask for.

How do I tell whether a problem wants me to prove a quadrilateral is a parallelogram or use the fact that it already is one

Read what is given versus what is asked. If the problem gives you side, angle, or diagonal information and asks you to "show" or "prove" the figure is a parallelogram, you need a converse — pick whichever one matches the given data (equal opposite sides, equal opposite angles, one pair equal-and-parallel, or bisecting diagonals). If the problem already states " $ABCD$ is a parallelogram" and asks you to find an angle, a side, or a diagonal length, you instead apply the forward properties directly, the way Worked Example 1 uses "opposite angles are equal" and "consecutive angles are supplementary" to solve for $x$ .