Here, \(\angle1=\angle3, \angle2=\angle4\ and\ \angle3=\angle4\), Euclid's first axiom says, the things which are equal to same things are equal to one another.

So, \(\angle 1=\angle 2\)

In the given figure,
PR = PQ   [radii ofthe same circle]    ... (i)
and QP = QR    [radii of the same circle] ... (ii) 
From Eqs. (i) and (ii), PR = QR   [by axiom 1] ... (iii) 
From Eqs. (i), (ii) and (iii), PQ = QR = PR   [by axiom 1]
Hence proved.

Given Three lines l, m and n, such that l || n and m II n.
To prove    l II m
Proof If possible, let l is not parallel to m, then I and m intersect in a unique point, say P. Thus, through a point P outside n, there are two lines I and m, both parallel to n. This is contradiction to the parallel line axiom. So, our supposition is wrong.
ஃ  l || m 
Hence proved.

Given, AB = BC    
Since M is the mid-point of AB.
ஃ  AM = MB = \(\frac{1}{2}\) AB
Also, N is the mid-point of BC
ஃ BN = NC = \(\frac{1}{2}\)BC
According to Euclid's axiom 7, things which are halves of the same things are equal to one another.
On multiplying both sides of Eq. (i) by\(\frac{1}{2}\), we get
\(\frac{1}{2}\)AB = \(\frac{1}{2}\)BC ⇒ AM = NC
[from Eqs. (ii) and (iii)]
(ii) Given, BM = BN
Since, M is the mid-point of AB.
ஃ AM = BM =\(\frac{1}{2}\)AB
⇒ 2AM = 2BM = AB
Also, N is the mid-point of BC.
ஃ BN = NC =\(\frac{1}{2}\)BC
⇒ 2BN = 2NC = BC       ...(iii)
According to Euclid's axiom 6, things which are double of the same things, are equal to one another.
On multiplying Eq. (i) by 2, we get
2BM = 2BN ⇒AB = BC [from Eqs. (ii) and (iii)]

(i) False. This can be seen visually.
(ii) False. This contradicts the Axiom.
[Given two distinct points, there is a unique line that passes through them.]
(iii) True by Euclid's Postulate 
[A terminated line can be produced indefinitely.]
(iv) True. If we superimpose the region bounded by one circle on the other, then they coincide. So, their centres and boundaries coincide, therefore, their radii will coincide.
(v) True by the first Axiom of Euclid.
[Things which are equal to the same thing are equal to one another.]

\(\angle \)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

AC = XD I Given
2AC = 2XD
\(\therefore \)  Things which are double of the same things are equal to one another
\(\Rightarrow \) AB = XY
\(\therefore \)  C is the midpoint of AB and D is the midpoint of XY

Given, a point C lies between two points A and B such that AC = BC.

On adding AC to both sides, we get
AC + AC = BC + AC ⇒ 2AC = AB
⇒ AC = \(\frac{1}{2}\) AB  Hence proved

(i) False, because from a single point, infinite number of lines can pass.

(ii) False, because from two distinct points, only one straight line can pass. [by postulate axiom]

(iii) True, it is Euclid's postulate 2.

(iv) True, because radii of congruent (equal) circles are always equal. In other words, if we superimpose the region bounded by one circle on the other circle, then they coincide. Then, their centres and boundaries also a>incide. Therefore, their radii will be same.

(v) True,given that AB = PQ (i) and PQ = XY (ii)
From Eqs. (i) and (ii), AB = XY

Here, \(OX=\frac{1}{2}XY, PX=\frac{1}{2}XZ\)
XY = 2(OX), XZ = 2(PX)
Also. OX = PX(Given)
XY = XZ
(because things which are double of the same things are equal to one another)

AC = DE(Given)
CB = CE
Adding, AC + CB = DC + CE
AB = DE
If equals are added to equals, the wholes are equal.

(i) False
(ii) False
(iii) True
(iv) False

Given, ㄥ2 = ㄥ1 and ㄥ3 = ㄥ4    ......(i)
Now,  ㄥ2 = ㄥ1
ㄥ2 + ㄥ3 = ㄥ1 + ㄥ3    [if equals are added to equals, then the wholes are equal ]
⇒ ㄥ2 + ㄥ3 = ㄥ2 + ㄥ3      [from Eq. (i)]
⇒ ㄥ2 + ㄥ3 + 30° = ㄥ2 + ㄥ3 + 30° [if equals are added to equals, then the wholes are equal]
⇒ ㄥ2 + ㄥ3 + 30° =ㄥ2 +ㄥ3 + θ ⇒ 30° = θ
[since, if equals are subtracted from equals, then the remainders are equal]
ஃ  θ = 30°

Given, QX = \(\frac{1}{2}\) XY       ...(i)
PX=\(\frac{1}{2}\)XZ        ....(ii)
and    QX= PX    ......(iii)
From Eqs. (i), (ii) and (iii), we get
\(\frac{1}{2}\)XY =\(\frac{1}{2}\)XZ
ஃ  XY = XZ     [by axiom 6]

(i) \(\overleftrightarrow { EF } ,\overleftrightarrow { GH } ,R\)
(ii) \(\overleftrightarrow { AB } ,\overleftrightarrow { EF } ,\overleftrightarrow { GH } ,R\)
(iii)  \(\overrightarrow { RB } ,\overrightarrow { RH } ,\overrightarrow { RG } \)
(iv)  \(\overline { RQ } ,\overline { RP } \)