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Parallel Lines

Definition

In 2-dimensional Euclidean geometry, two lines \( a\) and \( b\) are parallel if they do not intersect. We are familiar with several properties of parallel lines \( a\) and \( b\):

Property A

Given any point \( A\) on line \( a\), the minimum distance to line \( b\) is a constant. The converse is true; if the minimum distance between two lines is a constant, then they are parallel.

Property B

If line \( c\) intersects lines \( a\) and \( b\), then the corresponding angles of intersection are the same. The converse is true; if the corresponding angles of intersection are the same, then \( a\) and \( b\) are parallel.

Property C

If line \( c\) intersects lines \( a\) and \( b\), then the alternate interior angles are the same. The converse is true; if the corresponding alternate interior angles are the same, then \( a\) and \( b\) are parallel.

This is a useful property of parallel lines involving similar triangles.

Property D

If \( A, B\) are points on line \( a\), and \( C, D\) are points on line \( b\), let lines \( AC, BD\) intersect at \( E\). Then, triangles \( EAB\) and \( ECD\) are similar. The converse is also true. If \( EAC\) and \( EBD\) are straight lines, and triangles \( EAB, ECD\) are similar, then \( AB\) is parallel to \( CD\).

Proof: By property B, \( \angle EAB = \angle ACD, \angle EBA = \angle EDC\), so these 2 triangles have 3 corresponding angles equal. Thus, they are similar. (Students are asked to Test Yourself by proving the converse)

Remark: It does not matter if point \( E\) is between the two lines, or on the same side of both lines.

This is a useful property of parallel lines, involving area of a triangle.

Property E

If \( A, B\) are points on line \( a\), and \( C, D\) are points on line \( b\), then \( [ABC] = [ABD]\). (Note: \( [PQRS]\) represents the area of figure \( PQRS\).) The converse is also true.

Proof: The triangles have the same base \( AB\), and have the same height from property A. Thus, they have the same area. (Students are asked to Test Yourself by proving the converse.)

Technique

In a parallelogram, what is the sum of 2 consecutive angles?

Let \( \alpha\) and \( \beta\) be 2 consecutive angles in a parallologram. From property B, \( \alpha \) will be equal to the angle supplementary to \( \beta\), or that \( \alpha = 180^\circ - \beta\). Thus, the sum of 2 consecutive angles is \( 180^\circ\). (Pop quiz: Can you make a similar statement regarding trapeziums?)

 

\( AB\) and \( CD\) are 2 parallel lines. \( AC\) and \( BD\) intersect at \( E\). If \( EA=6, EB=10\) and \( AC=9\), what is the length \( BD?\)

From property D, we know that triangle \( EAB, ECD\) are similar. Hence, the ratio of their side lengths is the same. Thus \( \frac {EA}{EB} = \frac {EC}{ED} = \frac {AC}{BD}\), which allows us to calculate that \( BD = \frac {9 \cdot 10}{6} = 15\). Remark: Does it matter if point \( E\) is between the two lines, or on the same side of both lines? Draw both versions and compare?

 

\( AB\) and \( CD\) are 2 parallel lines. \( AC\) and \( BD\) intersect at \( E\). If \( [AED] = 11\), what is \( [BCE]\)?

From property E, \( [ABD] = [ABC]\). Then, we may add (or subtract, if \( E\) is between the two lines) triangle \( EAB\) to get that \( [BCE]=[AED]=11 \).

 

Point \( X\) is given between 2 parallel lines \( a\) and \( b\). The base of the perpendicular from \( X\) to \( a\) is denoted as \( A\), the base of the perpendicular from \( X\) to \( b\) is denoted as \( B\). Show that the points \( X, A, B\) lie on the same line.

Through \( X\), construct line \( x\) that is parallel to \( a\). Label another point \( Y\) on \( x\). Then, by Worked Example 1, \( 180^\circ = \angle AXY + 90^\circ\) and \( 180^\circ = \angle BXY + 90^\circ\). This gives us \( \angle AXY = 90 ^ \circ, \angle BXY = 90^\circ \). Thus, \( \angle AXY + \angle BXY = 180^\circ\), which shows that the points \( X, A, B\) lie on the same line.

Remark : Is this statement still true if \( X\) is on the same side of both lines?

Note by Arron Kau
3 years, 2 months ago

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