Since \(\angle1=\angle3\ and \ \angle2=\angle4\), therefore adding before equations.
\(\angle1+\angle2=\angle3+\angle4\)
\(\Rightarrow \angle BAD= \angle BCD\)
\(\Rightarrow \angle A= \angle C.\)

AB = BC (given)
BX = BY (given)
If equals are subtracted from equals, then remains are also equal.
AB - BX = BC - BY
\(\Rightarrow\) AX = CY

If a straight line I falls on two straight lines m and n such that sum of the interior angles on one side of I is two right angles, then by Euclid's fifth postulate the lines m and n will not meet on this side of I. Next, we know that the sum of the interior angles on the other side of line I will also be two right angles. Therefore, they will not meet on the other side also. So, the lines m and n never meet and are, therefore arallel. 

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

(i) Parallel lines. Lines which do not intersect anywhere are called parallel lines.
(ii) Perpendicular lines. Two lines which are at a right angle to each other are called perpendicular lines.
(iii) Line segment. It is a terminated line.
(iv) Radius. The length of the line segment joining the centre of a circle to any point on its circumference is called its radius.
(v) Square. A quadrilateral with all the four sides equal and all the four angles of measure 90° each is called a square.

(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.

AD is part of AF.
As whole is greater than part
Two more axioms
If equals are added two equals, the whole are equal.
e.g., if \(m\angle1=m\angle2,\) then
\(m\angle1+m\angle3=m\angle2+m\angle3\)
if equals are subtracted from equals, the remainders are equal.
e.g., if \(m\angle1=m\angle2\), then
\(m\angle1-m\angle3=m\angle2-m\angle3\) 

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
\(m\angle3=m\angle3\)
\(=m\angle2-m\angle3\)

AC = BC             I Given

⇒ AC + AC =  BC + AC         | If equals are added to equals, the wholes are equal.
⇒ 2AC = AB
AC = \(1\over2\)  AB
(ii) AC = OC              I Given
AC = \(1\over2\)AB                  I Proved in (i) above
\(\therefore\) CB = \(1\over2\)  AB
(iii) AC = BC                I Given
\(\therefore\) AC: BC = 1 : 1
\(\therefore\) Apala is correct. So, the value 'Sharpness' is depicted by her statement.
(iv) The mathematical concept 'Introduction to Euclid's Geometry' has been covered in this problem.
(v) The formulae used in the solution are as follows:
1. If equals are added to equals, the wholes are equal.
2. Concept of ratio.

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))] 

Let line segment \(\overline{AB}\) has 2 mid-points, say X and Y
then, \(\frac{AB}{2}=AX\ and\ \frac{AB}{2}=AY\)
\(\therefore\) AX = AY
X and Y coincides

AB = 2AE ( E is the mid-point of AB)
CD = 2DF (F is the mid-point of CD)
Also, AE = DF(Given)
Therefore, AB = CD(things which are double of the same things are equal to one another)

PQ = QR
QX = QY

If equals are subtracted from equals, the remainders are also equal.
We have PQ - QX = QR - QY
PX = RY

x + 4 = 10
x + 4 - 4 = 10 - 4
x = 6
If equals are subtracted from equals, the remainder are equal.

Things which are equal to the same thing are equal to one another.