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)

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)

Since, AB = BC
AX + BX = BY + CY
Since, BX = BY
AX + BX - BX = BY + CY - BY
AX = CY
Axiom : If equals are subtracted from the equals, the remainders are equal.

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

A system of axiom is called consistent if there is no statement which can be deduced from these axioms such that it contradicts any axiom. We know that if a transversal intersects two parallel lines, then each pair of corresponding angles are equal, which is a theorem. So, statement I is false hence, it is not an axiom.
Also, we know that if a transversal intersects two parallel lines, then each pair of alternate interior angles are equal. It is also a theorem. So, statement II is true, hence it is an axiom. Thus, in given statements, I is false and II is an axiom. Hence, given system of axioms is not consistent.

\(x+y=190^{ 0 }\neq 180^{ 0 }\) 

a =\(126^{ 0 }\), b =\(54^{ 0 }\), c =\(126^{ 0 }\), d =\(54^{ 0 }\)

Let \(\angle AOB=2x\) and \(\angle BOC=3x\)
\(\angle AOB+\angle BOC=\angle AOC\)
2x+3x=75^o
\(\Rightarrow x=\frac{75^o}{5}=15^o\)
\(\angle AOB=2x=30^o\) and \(\angle BOC=3x=45^o\)

\(90^{ 0 }\)

\(36^{ 0 },54^{ 0 },90^{ 0 }\)

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

x = 32.5°

Let ㄥA = 5x, ㄥB = 3x and ㄥC = 2x.
In ΔABC, we have ㄥA + ㄥB + ㄥC = 180°
⇒  5x+ 3x+ 2x = 180°
⇒   x=\(\frac{180^{\circ}}{10}\)=18°
ஃ ㄥA=5x=5x18°=90°
ㄥB=3x=3x18°​​​​​​​=54°​​​​​​​
and ㄥC = 2x = 2 x 18°​​​​​​​ = 36° 
Now, ㄥACD = ㄥBAC + ㄥABC 
[ஃ   exterior angle = sum of interior opposite angles]
⇒ ㄥACE + ㄥECD = 90° + 54°
ஃ ㄥECD = 54°[∵  ㄥACE = 90°] 

45°

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)]

Construction: Draw ray BL \(\bot \) PQ and ray CM \(\bot \) RS.

BL \(\bot \) PQ,CM \(\bot \) RS and PQ || RS BL||CM
\(\angle \) LBC=\(\angle \) MCB
|Alternate Interior Angles
\(\angle\)ABL=\(\angle\)LBC
Angle of incidence= Angle of reflection
\(\angle \)MCB=\(\angle \)MCD
Angle of incidence= Angle of reflection
From (1),(2) and (3) we get
\(\angle \)ABL =\(\angle \)MCD
Adding (1) and (4) , we get
\(\angle \) LBC+\(\angle \)ABL =\(\angle \)MCB+\(\angle \)MCD
\(\Rightarrow \) \(\angle \)ABC=\(\angle \)BCD
But these form a pair of equal alternate interior angles
So AB || CD.

AB||DC
\(\angle CDB=\angle ABD\)
=x=35^o[alternate angles]
x+y+80^o=180^o
\(\angle ADB=y=180^o-35^o-80^o\)
=65[angle sum property]
\(\angle DCB=z=180^o-[35^o+35^o]\)
=110^o

Given l || m
To prove  \(\angle 1+\angle 2-\angle 3=180^{ 0 }\)
Construction : Through C, draw CF || L || M

\(\therefore l||CF\) | by construction and a transversal BC intersects then
\(\therefore \angle 1+\angle FCB=180^{ 0 }\)
\(\therefore \)The Sum of consecutive interior angles on the same side of a transversal is \(180^{ 0 }\)
\(\Rightarrow \angle 1+\angle FCD=180^{ 0 }\)
BUT \(\angle FCD=\angle 3\)
| Alternate interior angles 
From (1) and (2)
\(\angle 1+\angle 2-\angle 3=180^{ 0 }\) 

\(\angle \)1=4x+15 | Corresponding angles
2x=180-\(\angle \)1 |Corresponding angles
\(\Rightarrow \) 2x=\(180^{ 0 }\) -(4x+15)
\(\Rightarrow \) 2x=165-4x
\(\Rightarrow \) 6x=165 \(\Rightarrow \)\(x=\frac { 165 }{ 6 } =27\frac { 1^{ 0 } }{ 2 } \)

Let \(l\bot m\)
Then \(\angle 1=90^{ 0 }\)
\(P\bot 1\)
Then \(\angle 2=90^{ 0 }\)

\(\therefore PS\bot L\) and RQ \(\bot\) L
PS || PQ
\(y^{ 2 }+\angle SRQ=180^{ 0 }\)

\(\angle PQS+\angle RSQ=180^{ 0 }\) 
\(\Rightarrow 60^{ 0 }+\angle RSQ=180^{ 0 }\) 
\(\Rightarrow \angle RSQ=120^{ 0 } \) 
\(\angle RFE=\angle RSQ=120^{ 0 }\)

\(\angle BC||DE\) and \(\angle CDE=90^{ 0 }\quad \)
\(\therefore \angle BCD=90^{ 0 }\) | Alternate Interior angles
\(\angle CDE=90^{ 0 }\) 
\(\therefore \angle DCE=180^{ 0 }-(30^{ 0 }+90^{ 0 })=60^{ 0 }\)

\(\angle a+\angle b=180^{ 0 }\) 
2 + 3 = 5
\(\therefore \angle b=\frac { 3 }{ 5 } x180^{ 0 }=108^{ 0 }=\angle h\)  

\(x+40^{ 0 }=180^{ 0 }\)

Corresponding angles axiom

The sum of consecutive interior angles on the same side of a transversal is \(180^{ 0 }\)

\(\angle 1\) and \(\angle 2\) are simply adjacent angles

\(\angle 2\) and \(\angle 8\) are alternate interior angles