Theorem 1 Two lines perpendicular to the same line are parallel.

Theorem 2 Parallel lines are equidistant everywhere.

Theorem 3 Two parallel straight lines are intercepted by a third straight line, and the interior angles are equal.

## Proof: Theorem 1

straight linel1⊥l3，l2⊥l3, to prove:l1l2.

Prove by contradiction. As shown in the figure:

Assumptionl1andl2not parallel, straightl3hand over separatelyl1，l2AtA，B, you can passBmake andl1parallel linesl4,AssumeP，Qrespectivelyl2，l4different from the pointB, ( the property theorem that parallel lines are perpendicular to straight lines ) thenl3andl4vertical, with∠ABQ=90∘. andl3andl2vertical, also∠ABP=90∘. This is the straight lineBPandBQdoes not coincide with the contradiction, and therefore assumesl1andl2Non-parallel is not established, so l1l2。

## Proof: Theorem 2

straight linel1l2，AB⊥l1，AB⊥l2，CD⊥l1，CD⊥l2, to prove:AB=CD.

Depend onl1l2, (by the theorem of the relationship between parallel lines and area ) , then

S△ABD=S△BDC

And ( the formula for the area of the sine of the angle between the two sides of the triangle )

S△ABD=12AB⋅BD⋅sin90∘=12AB⋅BD

S△BDC=12BD⋅CD⋅sin90∘=12BD⋅CD

therebyAB=CD

## Proof: Theorem 3

straight linel1l2, while the straight linel3hand over separatelyl1，l2AtA，B, Prove that the interior angles are equal.

Not fortifiedl3andl1，l2not vertical. PassA，Bdol1，l2vertical lines, respectivelyl2，l1AtP，Q,butPA⊥PB,then

S△BAPS△ABQ=S△PABS△APQ=PA⋅PBPA⋅QA=PBQA=PB⋅ABQA⋅AB

According to the inverse proportion theorem of common angle ,∠PBAand∠BAQequal or complementary, and both are acute angles, helping∠PBA=∠BAQ, that is, the interior angles are equal.

Note From this conclusion, it can be immediately deduced that two parallel lines are intercepted by a third straight line, the same angle is equal, and the same side interior angles are complementary.

### Replenish:

- By the parallel line and area relation theorem ,S△ABQ=S△APQ
- From the formula of the sine area of the angle between the two sides of the triangle , we getS△PAB=12PA⋅PB⋅sin∠APB=12PA⋅PB⋅sin90∘=12PA⋅PB, similarly