Solution

(i) False. Since through a single point, infinite number of lines can pass. In the following figure, it can be seen that there are infinite numbers of lines passing through a single point $P$.

(ii) False. Since through two distinct points, only one line can pass. In the following figure, it can be seen that there is only one single line that can pass through two

distinct points $P$ and $Q$.

(iii) True. A terminated line can be produced indefinitely on both the sides.

Let $AB$ be a terminated line. It can be seen that it can be produced indefinitely on both the sides.

(iv)True. If two circles are equal, then their centre and circumference will coincide and hence, the radii will also be equal.

(v) True. It is given that $AB$ and $XY$ are two terminated lines and both are equal to a third line PQ. Euclid’s first axiom states that things which are equal to the same thing are equal to one another. Therefore, the lines $AB$ and $XY$ will be equal to each other.

Solution

(i) False. Since through a single point, infinite number of lines can pass. In the following figure, it can be seen that there are infinite numbers of lines passing through a single point $P$.

(ii) False. Since through two distinct points, only one line can pass. In the following figure, it can be seen that there is only one single line that can pass through two

distinct points $P$ and $Q$.

(iii) True. A terminated line can be produced indefinitely on both the sides.

Let $AB$ be a terminated line. It can be seen that it can be produced indefinitely on both the sides.

(iv)True. If two circles are equal, then their centre and circumference will coincide and hence, the radii will also be equal.

(v) True. It is given that $AB$ and $XY$ are two terminated lines and both are equal to a third line PQ. Euclid’s first axiom states that things which are equal to the same thing are equal to one another. Therefore, the lines $AB$ and $XY$ will be equal to each other.

(iv) Radius of a circle

It is the distance between the centres of a circle to any point lying on the circle. To define the radius of a circle, we must know about point and circle.

(v) Square

A square is a quadrilateral having all sides of equal length and all angles of same measure, i.e., ${90}^{\circ }$. To define square, we must know about quadrilateral, side, and

angle.

Solution

There are various undefined terms in the given postulates.

The given postulates are consistent because they refer to two different situations. Also, it is impossible to deduce any statement that contradicts any well known axiom and postulate.

These postulates do not follow from Euclid’s postulates. They follow from the axiom, “Given two distinct points, there is a unique line that passes through them”.

Solution

It is given that,

$AC=BC$

$AC+AC=BC+AC$

(Equals are added on both sides) … (1) Here, (BC

$+AC$ ) coincides with $AB$. It is known that things which coincide with one another are equal to one another.

$\therefore BC+AC=AB\dots$

It is also known that things which are equal to the same thing are equal to one another. Therefore, from equations (1) and (2), we obtain

$AC+AC=AB$

$2AC=AB$

$\therefore AC=\frac{1}{2}AB$

Solution

Let there be two mid-points, C and D.

$AC=CB$

$AC+AC=BC+AC$ (Equals are added on both sides)

Here, $(BC+AC)$ coincides with $AB$. It is known that things which coincide with one another are equal to one another.

$\therefore BC+AC=AB\dots$

It is also known that things which are equal to the same thing are equal to one another.

Therefore, from equations (1) and (2), we obtain

$AC+AC=AB$

$\Rightarrow 2AC=AB$

Similarly, by taking $D$ as the mid-point of $AB$, it can be proved that

$2AD=AB$

From equation (3) and (4), we obtain

$2AC=2AD$ (Things which are equal to the same thing are equal to one another.) $\Rightarrow$ $AC=AD$ (Things which are double of the same things are equal to one another.) This is possible only when point $C$ and $D$ are representing a single point.

Hence, our assumption is wrong and there can be only one mid-point of a given line segment.

Solution

From the figure, it can be observed that $AC$

$=AB+BC$

$BD=BC+CD$

It is given that $AC=BDAB$

$+BC=BC+CD(1)$

According to Euclid’s axiom, when equals are subtracted from equals, the remainders are also equal.

Subtracting BC from equation (1), we obtain

$AB+BC-BC=BC+CD-BC$

$AB=CD$

Solution

Axiom 5 states that the whole is greater than the part. This axiom is known as a universal truth because it holds true in any field, and not just in the field of mathematics. Let us take two cases - one in the field of mathematics, and one other than that.

Case I

Let $t$ represent a whole quantity and only $a,b,c$ are parts of it.

$t=a+b+c$

Clearly, $t$ will be greater than all its parts $a,b$, and $c$.

Therefore, it is rightly said that the whole is greater than the part.

Case II

Let us consider the continent Asia. Then, let us consider a country India which belongs to Asia. India is a part of Asia and it can also be observed that Asia is greater than India. That is why we can say that the whole is greater than the part.

This is true for anything in any part of the world and is thus a universal truth.