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
\(m\angle1-m\angle3=m\angle2-m\angle3\) 

Here \(OX=\frac{1}{2}XY\)
\(PX=\frac{1}{2}XZ\)
Also OX = PX
\(\Rightarrow \frac{1}{2}XY=\frac{1}{2}XZ\)
Things equal to half of equals, are equal to one another.
Two other axioms:
Things coincide with one another are equal to one another.
e.g., If \(\overline{AB}\) coincide with \(\overline{XY}\), such that A falls on X and B falls on Y, then \(\overline {AB}=\overline{XY}\)
The whole is greater than the part
e.g., if \(m\angle1=m\angle2+m\angle3,\ then\ m\angle1>m\angle2\ and\ m\angle1>m\angle3\)

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

\(\because\)  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)

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

We have,
x - 15 = 25
\(\Rightarrow \)  x - I5 + 15 = 25 + 15  | If equals are added to equals, the wholesare equal (Euclid's Axiom (ii))
\(\Rightarrow \) x = 40

Let AB be perpendicular to a line l and AP is any other line segment.
In right \(\triangle ABP,\angle B>\angle P,(\therefore \angle B=90^o)\)
\(\ \Rightarrow AP>AB\ or\ AB\)

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

Euclid's axioms
(i) Things which are equal to the same thing are equal to one another.
(ii) If equals are added to equals, the wholes are equal.

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

In the given figure, AC coincides with AB + BC.
Also, Euclid's axiom 4 says that things which coincide with one another, are equal to one another. So, it can be deduced that
AB + BC = AC

Given, AB = (x+3), BC = 2xand AC = (4x-5) If B lies on AC, then
⇒   4x - 5 = x + 3 + 2x
[putting the values of AB, BC and AC]
⇒  4x - 5 = 3x + 3 
⇒  4x - 5 + 5 = 3x +3 +5   [adding 5 on both sides]
⇒   4x = 3x +8 
⇒   4x-3x = 3x +8 -3x
[subtracting 3x from both sides]
⇒  x=8
Hence, the value of x is 8.

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.

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. 

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