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 = CD(Given)
\(\Rightarrow\)AB + BC = BC + CD
\(\Rightarrow\)AC = BD
Euclid's axiom used: If equals are added to equals, the wholes are equal.

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)

Playfair's Axiom(Statement) : For every line l and for every point P not lying on l, there exists a unique line m passing through P and parallel to l. it is equivalent to Euclid's fifth postulate. 

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

(i) Infinite, if they are collinear.
(ii) Only one, if they are non-collinear points.

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.

PQ = QR
QX = QY

If equals are subtracted from equals, the remainders are also equal.
We have PQ - QX = QR - QY
PX = RY

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

Let x be the weight of Ram and Ravi each. On gaining 2 kg, weight of Ram and Ravi will be (x+2) kg each. According to Euclid's second axiom, when equals are added to equals, the wholes are equal. So, weights of Ram and Ravi are again equal.

Things which are equal to the same thing 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.

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.

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.

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.