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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

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