(i) The terms need to be defined are:
Polygon: A simple closed figure made up of three or more line segments.
Line segment: part of a line with two end points.
line: undefined term.
Point: undefied term.
Angle: A figure formed by two rays with a common initial point.
Ray: Part of a line with one end point.
Right angle: Angle whose measure is 90^o
Undefined terms used are: Line, point
Euclid's fourth postulate says that "all right angles are equal to one another." in a square, all angles are right angles, therefore all angles are equal (From Euclid's fourth postulate)
Three line segments are equal to fourth line segment.
Therefore, all the four sides of a square are equal( by Euclid's first axiom " things which are equal to one another")
(ii) Introduction to Euclid's geometry
(iii) Equality leads to democracy.

Here \(OX=\frac{1}{2}XY\)
\(PX=\frac{1}{2}XZ\)
Also OX = PX
\(\Rightarrow \frac{1}{2}XY=\frac{1}{2}XZ\)
Things equal to half of equals, are equal to one another.
Two other axioms:
Things coincide with one another are equal to one another.
e.g., If \(\overline{AB}\) coincide with \(\overline{XY}\), such that A falls on X and B falls on Y, then \(\overline {AB}=\overline{XY}\)
The whole is greater than the part
e.g., if \(m\angle1=m\angle2+m\angle3,\ then\ m\angle1>m\angle2\ and\ m\angle1>m\angle3\)

Since \(\angle1=\angle3 \ and \ \angle2=\angle4,\) therefore adding both equation
\(\angle1+\angle2=\angle3+\angle4\)
\(\Rightarrow \angle BAD=\angle BCD\)
\(\Rightarrow \angle A=\angle C\)
If equals are added to equal, the wholes are equal
Two more axioms:
Things which are equal to the same thing are equal to one another
e.g., if \(\overline {AB}=\overline{ PQ} \ and\ \overline {PQ}=\overline {XY}, then\ \angle{AB}=\angle{XY}\)
If equals are subtracted from equals, the remainders are equal.
e.g., if \(m\angle1=m\angle2\) then
\(m\angle3=m\angle3\)
\(=m\angle2-m\angle3\)

\(\because\)  C is the midpoint of AB
\(\therefore \)  AC = CB
AC + AC = CB + AC
| If equals are added to equals, then   the wholes are equal (Euclid's Axiom (ii))]
\(\Rightarrow \)  2AC =  AB      I CB + AC coincides with AB 
\(\Rightarrow \)  \(1\over2\)(2AC) = \(1\over2\) AB
| Things which are halves of the same thing are equal (Euclid's Axiom (vii»]
\(\Rightarrow \)  AC =\(1\over2\)AB
\(\Rightarrow \) \(1\over2\)AC  = \(1\over2\)(\(1\over2\)AB)
| Things which are halves of the same thing are equal to one another (Euclid's Axiom (vii))]
\(1\over2\)AC = \(1\over2\)AB
AD = \(1\over4\)AB
\(\because\) D is the mid-point of AC
\(\therefore \)AD = DC =\(1\over2\)AC (as above)

We have
AB = BC
\(\Rightarrow \) AB - BX = BC - BX
|If equals are subtracted from equals, the remainders are equal (Euclid's Axiom (iii))
AB - BX = BC - BY  \(|\)\( \because\)  BX = BY
AB - BX coincides with AX;
BC - BY coincides with CY
[Things which coincide with one another are equal to one another (Euclid's Axiom (iv))]

We have, AX=CY
AB=CD
\(\Rightarrow \) AB + BC = CD + BC
| If equals are added to equals, the wholes are equal (Euclid's Axiom (ii))
AC= BD
I AB + BC coincides with AC;
CD + BC coincides with BD
[Things which coincide with one another are equal to one another (Euclid's Axiom (iv))]