**Theorem: If a straight line intersects another two straight lines and if the corresponding angles are equal to each other.**

**General enunciation**: If a straight line intersects another two straight lines and if the corresponding angles are equal to each other.

**Particular enunciation:** Let the straight line PQ intersects AB and CD at the points R and S respectively, so that it becomes exterior ∠PRB = interior opposite ∠RSD.

It is required to prove that, AB and CD are parallel.

**Proof:** Since the ∠ARS =∠PRB (being vertically opposite angles)

And ∠ PRB =∠RSD (Given)

∴∠ARS =∠RSD. But they are alternate angles.

∴AB and CD are parallel. (Proved

**Theorem: The sum of the two interior angles in the same side of the bisector is equal to those two right angles then two lines are parallel.**

**General enunciation**: The sum of the two interior angles in the same side of the bisector is equal to those two right angles then two lines are parallel.

**Particular enunciation**: Let the straight line EF intersects the straight lines AB and CD at the points R and S respectively, so that in the some side of PQ,

interior ∠BRS + interior ∠RSD = two right angles

It is required to prove that, AB and CD are parallel.

**Proof:** Now, ∠ARS +∠BRS = two right angles

And ∠BRS + ∠RSD = two right angles (Given)

∴∠ARS +∠BRS =∠BRS + ∠RSD.

Now subtracting ∠BRS from both the sides we get, ∠ARS =∠RSD. But they are alternate angles.

∴AB and CD are parallel. **(**Proved).

Theorem: The straight lines which are parallel to the same straight line are parallel to one another.

**General enunciation**: The straight lines which are parallel to the same straight line are parallel to one another.

**Particular Enunciation**: Let the straight lines AB, CD be each parallel to the straight line EF.

It is required to prove that, AB and CD are parallel to one another.

**Construction**: Draw a straight line PQ cutting AB, CD and EF at the points X, Y and Z respectively.

**Proof**: Since AB and EF are parallel, PQ is their transversal, then

∠AXQ = the alternate ∠PZF

Again, Since CD and EF are parallel; PQ is their transversal, then

∠PYD = the corresponding ∠PZF

∴ ∠AXQ = ∠PYD.

∴ AB and CD are parallel to one another. (Proved)