Lesson Mid-line of a trapezoid is the locus of points equidistant from its bases

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Mid-line of a trapezoid is the locus of points equidistant from its bases

In this lesson you will learn that a mid-line of a trapezoid is the locus of points equidistant from its bases.

Theorem 1

Any point at the mid-line of a trapezoid is equidistant from the two parallel lines containing the bases of a trapezoid.

Proof

Let ABCD be a trapezoid with the bases AB and DC (Figure 1).
Let EF be the mid-line of the trapezoid ABCD and P be a point at the mid-line EF.
We need to prove that the point P is equidistant from the parallel lines AB and DC.

Let us draw the perpendiculars PQ and PR from the point P to the bases AB and DC
respectively, where Q and R are the intersection points of the perpendiculars with
the bases. Since the distance from the point to the straight line is the length of the
perpendicular drawn from the point to the line (see the lesson The distance from a
point to a straight line in a plane in this site), we need to prove that the segments
PQ and PR are congruent.

The segments PQ and PR lie in one straight line QR.
Consider the trapezoid AQRD and the segment EP.
This segment is the line segment drawn from the midpoint E of the lateral side AD.

Figure 1. To the Theorem 1

Besides of this, the segment EP is the part of the mid-line EF and therefore is parallel to the bases of the trapezoid ABCD.

Thus the segment EP is the line segment drawn from the midpoint E of the lateral side AD of the trapezoid AQRD and parallel to the bases AQ and DR of this trapezoid.
Hence, the segment EP intersects the other lateral side QR of the trapezoid AQRD in its midpoint, in accordance with the Theorem 2 of the lesson Trapezoids and their mid-lines under the current topic in this site. In other words, the point P is the midpoint of the segment QR, and the segments PQ and PR have the same length.

This is what has to be proved.

Theorem 2

If a point in a plane is equidistant from the two parallel lines containing the bases of a trapezoid, then the point belongs to the straight line containing the trapezoid's mid-line.

Proof

Let ABCD be a trapezoid with the bases AB and DC and the mid-line EF (Figure 2).
Let P be a point in a plane equidistant from the straight lines AB and DC.
We need to prove that the point P lies in the straight line containing the mid-line EF
of the trapezoid ABCD.

If the point P coincides with the point E or F then the Theorem statement is true.
Otherwise let us draw the perpendiculars PQ and PR from the point P to the straight
lines AB and DC respectively, where Q and R are the intersection points of the
perpendiculars with these straight lines.

Figure 2. To the Theorem 2

The perpendiculars PQ and PR lie in one straight line QR.
Consider the trapezoid AQRD with the lateral side QR and the point P in this side. Let us draw the straight segment EP connecting the points E and P.
Since the point P is equidistant from the straight lines AB and DC, the perpendiculars PQ and PR have the same length (see the lesson The distance from a point to a straight line in a plane under the topic Length, distance, coordinates, metric length of the section Geometry in this site).

It means that the point P is the midpoint of the lateral side QR of the trapezoid AQRD.
Hence, the segment EP is the mid-line of the trapezoid AQRD.
It implies that the segment EP is parallel to the bases AQ and DR in accordance with the property of a mid-line of a trapezoid (see the lesson Trapezoids and their mid-lines under the current topic in this site).
Thus we have two straight lines EF and EP passing through the point E and parallel to the bases AB and DC of the trapezoid ABCD. Hence, the straight line EF and EP coincide, and the point P lies on the straight line EF containing the mid-line of the trapezoid ABCD.

The Theorem 2 is proved.

Summary

1. Any point at the mid-line of a trapezoid is equidistant from the two parallel lines containing the bases of the trapezoid.
2. If a point in a plane is equidistant from the two parallel lines containing the bases of a trapezoid, then the point belongs to the straight line containing the trapezoid's mid-line.

Taken together, these two statements mean that the mid-line of a trapezoid is the locus of points equidistant from the two straight lines containing the bases of the trapezoid.

As the final part of the lesson, the problem below shows how the proved properties work.

Problem 1

In a trapezoid, any straight line segment connecting a point at the shorter base with a point at the larger base is bisected by the mid-line of the trapezoid. Prove.

Proof

Let ABCD be a trapezoid with the bases AB and DC and the mid-line EF (Figure 3).
Let IJ be an arbitrary straight segment that have one endpoint I on the straight
line containing the base AB of the trapezoid and the other endpoint J on the straight
line containing the other base DC. We need to prove that the mid-line EF of the
trapezoid bisects the segment IJ at the intersection point G.

First of all, the mid-line EF really does intersect the segment IJ, since its endpoints
I and J lie in two different half-planes relative to the mid-line EF.

Keeping this in mind, let us draw the perpendicular segments GK and GL from the
intersection point G to the bases AB and DC correspondingly,

Figure 3. To the Problem 1

It is clear that the segments GK and GL lie in one straight line KL. Let us consider and compare the triangles GKI and GLJ.

These triangles are right-angled triangles.
They have the congruent legs GK and GL, since the point G belongs to the trapezod's mid-line and, therefore, is equidistant from the straight lines AB and DC.
The triangles GKI and GLJ have congruent acute angles IGK and JGL. These angles are congruent as the vertical angles (see the lesson Vertical angles under the topic Angles, complementary, supplementary angles of the section Geometry in this site).

Hence, the triangles GKI and GLJ are congruent in accordance with the ASA-test for the triangles congruency (see the lesson Congruence tests for triangles under the topic Triangles of the section Geometry in this site). Therefore, the segments IG and GJ are congruent as the corresponding sides (hypotenuses) in these triangles.

This is what had to be proved.

My other lessons on trapezoids in this site are
    -Trapezoids and their base angles,
    -Trapezoids and their mid-lines,
    -Diagonals of an isosceles trapezoid are congruent,
    -Solving problems on trapezoids,
    -Solving problems on isosceles trapezoids,
    -Trapezoid is uniquely defined by the lengths of its sides,
    -HOW TO construct a trapezoid given by the lengths of its sides   and
    -PROPERTIES OF TRAPEZOIDS
under the current topic,   and
    -Solved problems on trapezoids   and
    -Solved problems on isosceles trapezoids
under the topic Geometry of the section Word problems.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link GEOMETRY - YOUR ONLINE TEXTBOOK.

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