Class 9 Maths Chapter 12: Quadrilaterals — Notes, Examples & Practice

A quadrilateral ABCD has four vertices, four sides (AB, BC, CD, DA) and two diagonals (AC and BD). Sides that share a vertex are adjacent; sides that do not are opposite.

It is convex if every interior angle is less than $180^\circ$ (both diagonals lie inside); otherwise it is concave (one angle is reflex and one diagonal lies outside).

A diagonal splits a quadrilateral into two triangles, each with angle sum $180^\circ$. So the **interior angles of a quadrilateral add up to $360^\circ$**.

A parallelogram is a quadrilateral with both pairs of opposite sides parallel. Draw diagonal AC of parallelogram ABCD: alternate angles give $\angle BAC = \angle DCA$ and $\angle BCA = \angle DAC$, and AC is common, so $\triangle ABC \cong \triangle CDA$ (ASA).

From this congruence: opposite sides are equal ($AB = CD$, $BC = DA$) and opposite angles are equal. Adjacent angles are co-interior angles, so they add up to $180^\circ$.

Using the other diagonal as well, triangles AOB and COD are congruent (ASA), so the diagonals bisect each other: $AO = OC$ and $BO = OD$.

Most properties have true converses (Chapter 9), and each gives a test. A quadrilateral is a parallelogram if: both pairs of opposite sides are equal; or both pairs of opposite angles are equal; or the diagonals bisect each other.

The handiest test: one pair of opposite sides is both equal and parallel. If $AB = CD$ and $AB \parallel CD$, then triangles ABC and CDA are congruent (SAS), giving $BC \parallel AD$ too.

Careful: one pair of sides parallel and the other pair equal is not enough — an isosceles trapezium has that.

Midpoint theorem: the segment joining the midpoints of two sides of a triangle is parallel to the third side and half of it. If E and F are the midpoints of AB and AC, then $EF \parallel BC$ and $EF = \dfrac{1}{2}BC$.

Proof idea: extend EF to G so that $FG = EF$. Triangles AEF and CGF are congruent (SAS), so $CG = AE = EB$ and $CG \parallel AB$. Then EBCG has a pair of opposite sides equal and parallel, so it is a parallelogram, giving $EG \parallel BC$ and $EG = BC$, i.e. $EF = \dfrac{1}{2}BC$.

Converse: a line through the midpoint of one side, parallel to a second side, bisects the third side.

A median joins a vertex to the midpoint of the opposite side. A triangle's three medians always pass through one point, the centroid G — its balance point.

The centroid divides each median in the ratio 2 : 1, measured from the vertex. You can see why with the midpoint theorem: for medians BE and CF meeting at G, let M and N be the midpoints of BG and CG. Then FE and MN are both parallel to BC and equal to half of it, so FENM is a parallelogram; its diagonals bisect each other, giving $BM = MG = GE$, so $BG : GE = 2 : 1$.

In coordinates (Chapter 1), the centroid of $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is the average of the vertices.

Varignon's theorem: join the midpoints of the sides of any quadrilateral in order and you always get a parallelogram. Draw diagonal AC: by the midpoint theorem, the two midpoint segments on either side of it are both parallel to AC and half its length — one pair of opposite sides equal and parallel.

The shape of that parallelogram depends on the diagonals: equal diagonals give a rhombus, perpendicular diagonals give a rectangle, both give a square.

Tiling means covering the plane with copies of a shape with no gaps or overlaps. Any quadrilateral — even a concave one — tiles the plane: rotate a copy by $180^\circ$ about the midpoint of a side, and repeat. Around each vertex the four different angles meet, adding to exactly $360^\circ$.