Equidistance Property Of Parallel Lines

Theorem: Perpendicular distance between the two parallel lines is constant.

Prerequisites:
Interior angle property of parallel lines (proof)
Alternate angle property (proof)
AAS congruency (proof)

Proof:

Let $l_1$ and $l_2$ be two parallel lines. Let AC and BD be two perpendicular drawn from $l_1$ to $l_2$.

Since, for parallel lines, any two consecutive interior angles are supplementary, hence all the four angles in the figure, i.e., $\angle A,\; \angle B, \;\angle C\; \text{and}\; \angle D$ are $90^o$.

Now, consider $\triangle ABC$ and $\triangle DCB$:
$\qquad\quad\angle A = \angle D\qquad\qquad\qquad\qquad\qquad\:\!\text{(Right angles)}\\
\qquad\quad\angle ABC = \angle BCD\qquad\quad\qquad\qquad\text{(Alternate angles)}\\
\qquad\quad BC = BC\qquad\qquad\qquad\qquad\qquad\text{(Common)}$

Hence, $\triangle ABC\cong\triangle DCB$ by AAS rule.

Thus, by CPCTC, AC = BD.

Q.E.D.


Recommended:
Unique parallel through a point
Corresponding angles property
Unique RHS triangle