Parallel Lines Property Theorem

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 line$l_1\perp l_3$l1⊥l3，$l_2\perp l_3$l2⊥l3, to prove:$l_1{l}_2$l1l2.

Prove by contradiction. As shown in the figure:

Assumption$l_1$l1and$l_2$l2not parallel, straight$l_3$l3hand over separately$l_1$l1，$l_2$l2At$A$A，$B$B, you can pass$B$Bmake and$l_1$l1parallel lines$l_4$l4,Assume$P$P，$Q$Qrespectively$l_2$l2，$l_4$l4different from the point$B$B, ( the property theorem that parallel lines are perpendicular to straight lines ) then$l_3$l3and$l_4$l4vertical, with$\angle ABQ={90}^{\circ }$∠ABQ=90∘. and$l_3$l3and$l_2$l2vertical, also$\angle ABP={90}^{\circ }$∠ABP=90∘. This is the straight line$BP$BPand$BQ$BQdoes not coincide with the contradiction, and therefore assumes$l_1$l1and$l_2$l2Non-parallel is not established, so $l_1{l}_2$l1l2。

Proof: Theorem 2

straight line$l_1{l}_2$l1l2，$AB\perp l_1$AB⊥l1，$AB\perp l_2$AB⊥l2，$CD\perp l_1$CD⊥l1，$CD\perp l_2$CD⊥l2, to prove:$AB=CD$AB=CD.

Depend on$l_1{l}_2$l1l2, (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 )

$S_{△ABD}=\frac{1}{2}AB\cdot BD\cdot \sin {90}^{\circ }=\frac{1}{2}AB\cdot BD$S△ABD=12AB⋅BD⋅sin⁡90∘=12AB⋅BD

$S_{△BDC}=\frac{1}{2}BD\cdot CD\cdot \sin {90}^{\circ }=\frac{1}{2}BD\cdot CD$S△BDC=12BD⋅CD⋅sin⁡90∘=12BD⋅CD

thereby$AB=CD$AB=CD

Proof: Theorem 3

straight line$l_1{l}_2$l1l2, while the straight line$l_3$l3hand over separately$l_1$l1，$l_2$l2At$A$A，$B$B, Prove that the interior angles are equal.

Not fortified$l_3$l3and$l_1$l1，$l_2$l2not vertical. Pass$A$A，$B$Bdo$l_1$l1，$l_2$l2vertical lines, respectively$l_2$l2，$l_1$l1At$P$P，$Q$Q,but$PA\perp PB$PA⊥PB,then

$$\frac{S_{△BAP}}{S_{△ABQ}}=\frac{S_{△PAB}}{S_{△APQ}}=\frac{PA\cdot PB}{PA\cdot QA}=\frac{PB}{QA}=\frac{PB\cdot AB}{QA\cdot AB}$$

According to the inverse proportion theorem of common angle ,$\angle PBA$∠PBAand$\angle BAQ$∠BAQequal or complementary, and both are acute angles, helping$\angle PBA=\angle BAQ$∠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 ,$S_{△ABQ}=S_{△APQ}$S△ABQ=S△APQ
• From the formula of the sine area of ​​the angle between the two sides of the triangle , we get$S_{△PAB}=\frac{1}{2}PA\cdot PB\cdot \sin \angle APB=\frac{1}{2}PA\cdot PB\cdot \sin {90}^{\circ }=\frac{1}{2}PA\cdot PB$S△PAB=12PA⋅PB⋅sin⁡∠APB=12PA⋅PB⋅sin⁡90∘=12PA⋅PB, similarly$S_{△APQ}=\frac{1}{2}PA\cdot QA$