Construction of Quadrilaterals

A quadrilateral is a four-sided polygon enclosed on a flat plane by four distinct straight line segments meeting at four vertices. Connecting any two non-consecutive vertices creates an internal line segment called a diagonal. Every simple quadrilateral has exactly four vertices, four sides, four interior angles, and two diagonals.

Theorem: The sum of the interior angles of any convex quadrilateral is always four right angles, or $360^\circ$. Symbolically, $\angle A + \angle B + \angle C + \angle D = 360^\circ$.

Let $ABCD$ be a convex quadrilateral. Draw diagonal $AC$. This diagonal dissects the quadrilateral into two non-overlapping triangles: $\triangle ABC$ and $\triangle ADC$.
From triangle geometry, the sum of angles in $\triangle ABC$ is $180^\circ$, and in $\triangle ADC$ is $180^\circ$.
Adding both systems: $(\angle CAB + \angle DAC) + \angle B + (\angle BCA + \angle DCA) + \angle D = 180^\circ + 180^\circ = 360^\circ$.
Since $\angle CAB + \angle DAC = \angle A$ and $\angle BCA + \angle DCA = \angle C$, we conclude $\angle A + \angle B + \angle C + \angle D = 360^\circ$.

If four angles of a quadrilateral are in the ratio $1:2:3:4$, what is the measure of each angle?
Let the angles be $x, 2x, 3x, 4x$.
$x + 2x + 3x + 4x = 360^\circ \implies 10x = 360^\circ \implies x = 36^\circ$.
The angles are $36^\circ, 72^\circ, 108^\circ, 144^\circ$.

Exam Trap: Never confuse the polygon angle sum with $180^\circ$. Triangles sum to $180^\circ$; quadrilaterals sum to $360^\circ$. Furthermore, no quadrilateral can possess four obtuse angles ($>90^\circ$), as their sum would strictly exceed $360^\circ$.

A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel ($AB \parallel CD$ and $AD \parallel BC$). This dual parallelism generates remarkable symmetry across sides, angles, and diagonals.

Key Invariant Properties of a Parallelogram:
1. Opposite sides are congruent ($AB = CD$ and $AD = BC$).
2. Opposite angles are congruent ($\angle A = \angle C$ and \angle B = \angle D$).
3. Any two consecutive adjacent angles are supplementary ($\angle A + \angle B = 180^\circ$).
4. Diagonals bisect each other ($OA = OC$ and $OB = OD$).

Because $AB \parallel CD$ and transversal $BC$ cuts them, consecutive interior angles must sum to $180^\circ$: $\angle B + \angle C = 180^\circ$. Drawing diagonal $AC$ creates congruent triangles $\triangle ABC \cong \triangle CDA$ by ASA criterion, proving opposite sides and opposite angles are identically equal.

If one angle of a parallelogram is $72^\circ$, find all other angles.
Opposite angle $= 72^\circ$. Adjacent angles $= 180^\circ - 72^\circ = 108^\circ$.
The four angles are $72^\circ, 108^\circ, 72^\circ, 108^\circ$.

Sufficient Condition: If just ONE pair of opposite sides of a quadrilateral is BOTH equal and parallel, the figure is guaranteed to be a parallelogram!

Rectangle: A parallelogram containing at least one right angle ($90^\circ$). By angle sum properties, all four interior angles become $90^\circ$.
Square: A regular quadrilateral that is simultaneously a rectangle and a rhombus—all four sides are equal and all four angles measure $90^\circ$.

Diagonal Invariants:
- Rectangle: Diagonals are equal in length ($AC = BD$) and bisect each other.
- Square: Diagonals are equal in length ($AC = BD$), bisect each other, AND intersect at right angles ($AC \perp BD$ at $90^\circ$). Length of diagonal for side $a$ is $d = a\sqrt{2}$.

In rectangle $ABCD$, consider right triangles $\triangle ABC$ and $\triangle DCB$. $AB = DC$ (opposite sides), $\angle B = \angle C = 90^\circ$, and $BC$ is common. By SAS congruence, $\triangle ABC \cong \triangle DCB$, directly implying hypotenuses $AC = BD$.

A rectangular plot has length $12\text{ m}$ and width $5\text{ m}$. Find the diagonal distance.
$d = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ m}$.

Logical Inclusion: "Every square is a rectangle, but not every rectangle is a square." A rectangle only becomes a square when its adjacent sides are equal.

Rhombus: An equilateral parallelogram having all four sides of equal length. Its opposite sides remain parallel, and opposite angles are equal.
Kite: A quadrilateral with two distinct pairs of adjacent equal-length sides. It is not generally a parallelogram because its opposite sides are not parallel.

Perpendicular Diagonal Property:
The diagonals of a rhombus bisect each other perpendicularly ($AC \perp BD$ at $90^\circ$).
In a kite, the longer diagonal is the perpendicular bisector of the shorter diagonal.
Area of Rhombus or Kite $= \frac{1}{2} \times d_1 \times d_2$.

In rhombus $ABCD$ with center $O$, triangles $\triangle AOB$ and $\triangle COB$ have $AB = BC$ (equal sides), $OA = OC$ (bisection), and $OB$ common. By SSS congruence, $\triangle AOB \cong \triangle COB$. Thus $\angle AOB = \angle COB$. Since they form a linear pair, each angle must be $\frac{180^\circ}{2} = 90^\circ$.

Find the side of a rhombus whose diagonals are $10\text{ cm}$ and $24\text{ cm}$.
Semi-diagonals are $5\text{ cm}$ and $12\text{ cm}$.
Side $= \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ cm}$.

Frequent Fallacy: Rhombus diagonals are NOT equal unless the rhombus is a square! Never set $d_1 = d_2$ for a non-square rhombus.

A trapezium (or trapezoid) is a quadrilateral with at least one pair of parallel opposite sides ($AB \parallel CD$). The non-parallel sides are called legs.
If the two non-parallel legs are equal in length ($AD = BC$), it is termed an isosceles trapezium.

Isosceles Trapezium Invariants:
1. Base angles on the same parallel side are equal ($\angle A = \angle B$ and $\angle D = \angle C$).
2. Both diagonals are equal in length ($AC = BD$).
3. Consecutive interior angles along non-parallel sides are supplementary ($\angle A + \angle D = 180^\circ$).

Because $AB \parallel CD$, the transversal line along leg $AD$ produces consecutive interior angles satisfying $\angle A + \angle D = 180^\circ$. In an isosceles trapezium, dropping perpendicular heights from $D$ and $C$ onto $AB$ creates two congruent right triangles, proving base angles and diagonals are identically equal.

In an isosceles trapezium, one base angle is $60^\circ$. Find the remaining three angles.
Base angles are equal $\implies \angle B = 60^\circ$.
Supplementary upper angles $\implies 180^\circ - 60^\circ = 120^\circ$.
The four angles are $60^\circ, 60^\circ, 120^\circ, 120^\circ$.

Exam Trap: A general trapezium does NOT have equal diagonals. Only when the word "isosceles" is specified do the diagonals become equal.