Consider walking around the polygon along its perimeter. At each vertex you turn through the exterior angle to continue to the next side. After returning to the start you will have turned through exactly one full revolution, which is \(360^\circ\). Therefore, the sum of the exterior angles = \(360^\circ\). This reasoning holds for any polygon irrespective of number of sides or whether it is convex or concave (for concave polygons, take exterior angle as the exterior turn's measure).

Let \(ABCD\) be a parallelogram (so \(AB\parallel CD\) and \(AD\parallel BC\)).

(i) Draw diagonal \(AC\). Triangles \(ABC\) and \(CDA\) have \(\angle CAB = \angle ACD\) (alternate interior angles) and \(\angle CBA = \angle CAD\). AC is common. So by ASA, \(\triangle ABC \cong \triangle CDA\) ⇒ \(AB = CD\) and \(BC = AD\).

(ii) From the same congruency, corresponding angles are equal, so \(\angle A = \angle C\) and \(\angle B = \angle D\).

(iii) Let diagonals \(AC\) and \(BD\) meet at \(O\). From congruent triangles formed by the diagonals (\(\triangle AOB \cong \triangle COD\) by ASA), we get \(AO = OC\) and \(BO = OD\). Thus diagonals bisect each other at \(O\).

Let \(ABCD\) be a rhombus (so \(AB = BC = CD = DA\)). Diagonals intersect at \(O\). Consider triangles \(AOB\) and \(COB\).

We have \(AB = CB\) (sides of rhombus), \(OB\) common and \(AO = OC\) ? — To show bisector property we proceed differently: using triangles \(AOB\) and \(COD\), or use symmetry/folding: a rhombus is a parallelogram so diagonals bisect each other. Also since adjacent sides equal, triangles formed by halves show that the diagonals are perpendicular: prove using SSS that triangles around intersection are congruent and the angles formed are right angles — thus diagonals perpendicular. (One formal way: show \(\triangle AOB \cong \triangle BOC\) etc. then use linear pair argument to get each of the four angles at \(O\) are equal; as they sum to \(360^\circ\), each is \(90^\circ\).)

Let rectangle \(ABCD\) with right angles. Consider triangles \(ABC\) and \(CDA\). AB = CD (opposite sides equal), BC = AD and \(\angle B = \angle D = 90^\circ\). By SAS, triangles are congruent, thus diagonals \(AC = BD\). Another standard proof: in right triangles formed by diagonal, use Pythagoras on both and show equal results.

Following the example style: If \(OE=4\), then \(OP=4\) (diagonals bisect). So \(PE = 8\). Then \(HL = PE + 5 = 8 + 5 = 13\). \(OH = \dfrac{1}{2}\times HL = \dfrac{13}{2} = 6.5\) cm.

Opposite angle \(\angle N = 70^\circ\). Adjacent angles: \(\angle I = \angle G = 180^\circ - 70^\circ = 110^\circ\). So angles are \(70^\circ,110^\circ,70^\circ,110^\circ\).

A square has equal sides and each angle \(90^\circ\). So it meets the definition of a rhombus (all sides equal) and a rectangle (all angles right). From these, the square inherits diagonals that are equal (rectangle property) and diagonals that are perpendicular bisectors (rhombus property). Thus it combines both sets of properties.

This is the parallelogram law. One standard vector/coordinate proof: Put \(A\) at origin, \(B\) at vector \(\mathbf{u}\), \(D\) at \(\mathbf{v}\). Then sides are \(\mathbf{u}\), \(\mathbf{v}\), \(\mathbf{u}\), \(\mathbf{v}\) and diagonals \(\mathbf{u}+\mathbf{v}\) and \(\mathbf{u}-\mathbf{v}\). Compute squared norms: \(2(\|\mathbf{u}\|^2 + \|\mathbf{v}\|^2) = \|\mathbf{u}+\mathbf{v}\|^2 + \|\mathbf{u}-\mathbf{v}\|^2\). Expanding gives the required relation.

• Parallelogram: diagonals bisect each other; not necessarily perpendicular; not necessarily equal.
• Rhombus: diagonals bisect each other and are perpendicular; not necessarily equal.
• Rectangle: diagonals bisect each other and are equal; not perpendicular necessarily.
• Square: diagonals bisect each other, are equal and are perpendicular.
• Kite: one diagonal typically bisects the other and diagonals are often perpendicular (but depends on kite).

If a trapezium has all sides equal, then both pairs of opposite sides would be equal too (since all sides equal) which forces both pairs to be parallel — that would make it a parallelogram. Combining equal sides and a pair of parallel sides often gives a square or rhombus shape; but for a trapezium definition (at least one pair of parallel sides) to hold with all sides equal, it would actually be a parallelogram (rhombus). Thus a trapezium having all sides equal must be a special parallelogram (rhombus), not a non-parallelogram trapezium.

Given adjacent angles equal, and knowing adjacent angles of parallelogram are supplementary if the quadrilateral is a parallelogram — but we need more conditions. (This problem needs precise conditions: If in a parallelogram two adjacent angles are equal, then each must be \(90^\circ\) making it a rectangle. Explanation: let \(\angle A = \angle B\) and \(\angle A + \angle B = 180^\circ\), so \(2\angle A = 180^\circ\) ⇒ \(\angle A = 90^\circ\). Thus rectangle.)

Square is a rhombus (all sides equal) and a rectangle (all angles \(90^\circ\)). From rectangle property diagonals are equal; from rhombus property diagonals are perpendicular. Combining both gives diagonals equal and perpendicular.

Construction: From point A draw AB of given length; at A construct included angle; from B draw a segment equal to the second side; from point created draw a line parallel to AB and from A draw a line parallel to BC to meet at D; connect to form ABCD. Proof uses alternate interior angles and congruency of triangles formed by a diagonal to show opposite sides equal.

Paper folding is a visual proof technique: fold along diagonals to check bisectors, fold edges to test symmetry (square and rhombus), fold along midlines to find perpendicularity (rhombus diagonals), and fold corners to verify equality of opposite sides or angles. These physical tests give intuition before formal proofs via congruency.