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 $l_{1} ⊥ l_{3}$ ， $l_{2} ⊥ l_{3}$ , to prove: $l_{1} l_{2}$ .

Prove by contradiction. As shown in the figure:

Assumption $l_{1}$ and $l_{2}$ not parallel, straight $l_{3}$ hand over separately $l_{1}$ ， $l_{2}$ At $A$ ， $B$ , you can pass $B$ make and $l_{1}$ parallel lines $l_{4}$ ,Assume $P$ ， $Q$ respectively $l_{2}$ ， $l_{4}$ different from the point $B$ , ( the property theorem that parallel lines are perpendicular to straight lines ) then $l_{3}$ and $l_{4}$ vertical, with $∠ A B Q = 90^{\circ}$ . and $l_{3}$ and $l_{2}$ vertical, also $∠ A B P = 90^{\circ}$ . This is the straight line $B P$ and $B Q$ does not coincide with the contradiction, and therefore assumes $l_{1}$ and $l_{2}$ Non-parallel is not established, so $l_{1} l_{2}$ 。

Proof: Theorem 2

straight line $l_{1} l_{2}$ ， $A B ⊥ l_{1}$ ， $A B ⊥ l_{2}$ ， $C D ⊥ l_{1}$ ， $C D ⊥ l_{2}$ , to prove: $A B = C D$ .

Depend on $l_{1} l_{2}$ , (by the theorem of the relationship between parallel lines and area ) , then

$S_{△ A B D} = S_{△ B D C}$

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

$S_{△ A B D} = \frac{1}{2} A B \cdot B D \cdot \sin 90^{\circ} = \frac{1}{2} A B \cdot B D$

$S_{△ B D C} = \frac{1}{2} B D \cdot C D \cdot \sin 90^{\circ} = \frac{1}{2} B D \cdot C D$

thereby $A B = C D$

Proof: Theorem 3

straight line $l_{1} l_{2}$ , while the straight line $l_{3}$ hand over separately $l_{1}$ ， $l_{2}$ At $A$ ， $B$ , Prove that the interior angles are equal.

Not fortified $l_{3}$ and $l_{1}$ ， $l_{2}$ not vertical. Pass $A$ ， $B$ do $l_{1}$ ， $l_{2}$ vertical lines, respectively $l_{2}$ ， $l_{1}$ At $P$ ， $Q$ ,but $P A ⊥ P B$ ,then

$\frac{S_{△ B A P}}{S_{△ A B Q}} = \frac{S_{△ P A B}}{S_{△ A P Q}} = \frac{P A \cdot P B}{P A \cdot Q A} = \frac{P B}{Q A} = \frac{P B \cdot A B}{Q A \cdot A B}$

According to the inverse proportion theorem of common angle , $∠ P B A$ and $∠ B A Q$ equal or complementary, and both are acute angles, helping $∠ P B A = ∠ B A Q$ , 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 , $S_{△ A B Q} = S_{△ A P Q}$
From the formula of the sine area of the angle between the two sides of the triangle , we get $S_{△ P A B} = \frac{1}{2} P A \cdot P B \cdot \sin ∠ A P B = \frac{1}{2} P A \cdot P B \cdot \sin 90^{\circ} = \frac{1}{2} P A \cdot P B$ , similarly $S_{△ A P Q} = \frac{1}{2} P A \cdot Q A$