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. 

Let a line AB have two mid-points, say, C and D. Then,
AC = \(1\over2\) AB     ...(1)
AD = \(1\over2\)AB     ...(2)
From (1) and (2),
AC = AD.
Things which are equal to the same thing are equal to one another

AC= BC
AC + AC = BC + AC         | Equals are added to equals  
\(\Rightarrow \) 2AC = AB                    I BC + AC coincides with AB
\(\Rightarrow \)   AC= \(1\over2\)2AB .
\(\Rightarrow \)  AC=BC= \(1\over2\)AB.
|Things which are equal to the same thing are equal to one another.

(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.

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.

\(\because\)  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)

\(\angle \)2 = \(\angle \)4          ,,,,,,,(1) | Given
\(\angle \)1= \(\angle \)4          ........(2) | Given
\(\angle \)3 =\(\angle \)2         .......(3) | Given
From (1), \(\angle \)4 = \(\angle \)2           ......(4)
From (3) and (4),
\(\angle \)3 = \(\angle \)4               .....(5)
I Things which are equal to the same thing are equal to one another
From (2) and (4),
\(\angle \)1 = \(\angle \)3
I Things which are equal to the same thing are equal to one another

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))] 

Take any line l and a point P not on l. Then, by Playfair’s axiom, which is equivalent to the fifth postulate, we know that there is a unique line m through P which is parallel to l.