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

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<H2>Mid-line of a trapezoid is the locus of points equidistant from its bases</H2>

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

<H3>Theorem 1</H3>Any point at the mid-line of a trapezoid is equidistant from the two parallel lines containing the bases of a trapezoid.

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Proof

Let &nbsp;ABCD&nbsp; be a trapezoid with the bases &nbsp;AB&nbsp; and &nbsp;DC&nbsp; (Figure 1).

Let &nbsp;EF&nbsp; be the mid-line of the trapezoid &nbsp;ABCD&nbsp; and &nbsp;P&nbsp; be a point at the mid-line &nbsp;EF.

We need to prove that the point &nbsp;P&nbsp; is equidistant from the parallel lines &nbsp;AB&nbsp; and &nbsp;DC.

Let us draw the perpendiculars &nbsp;PQ&nbsp; and &nbsp;PR&nbsp; from the point &nbsp;P&nbsp; to the bases &nbsp;AB&nbsp; and &nbsp;DC

respectively, &nbsp;where &nbsp;Q&nbsp; and &nbsp;R&nbsp; are the intersection points of the perpendiculars with

the bases. &nbsp;Since the distance from the point to the straight line is the length of the

perpendicular drawn from the point to the line &nbsp;(see the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Length-and-distance/The-distance-from-a-point-to-a-straight-line.lesson>The distance from a 
point to a straight line in a plane</A>&nbsp; in this site), &nbsp;we need to prove that the segments

PQ&nbsp; and &nbsp;PR&nbsp; are congruent.

The segments &nbsp;PQ&nbsp; and &nbsp;PR&nbsp; lie in one straight line &nbsp;QR.

Consider the trapezoid &nbsp;AQRD&nbsp; and the segment &nbsp;EP.

This segment is the line segment drawn from the midpoint &nbsp;E&nbsp; of the lateral side &nbsp;AD.

</TD>
 <TD>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{drawing( 280, 160, 0.5, 7.5, 0.5, 4.5, 
 line( 1, 1, 7, 1), 
 line( 1, 1, 3, 4),
 line( 3, 4, 6, 4), 
 line( 7, 1, 6, 4),

locate ( 0.9, 1,   A),
            locate ( 6.9, 1,   B),
            locate ( 6.1, 4.3, C),
            locate ( 2.7, 4.3, D),

line( 2.0, 2.5, 6.5, 2.5),
            locate ( 1.7, 2.7, E),
            locate ( 6.6, 2.7, F),

circle ( 4, 2.5, 0.07, 0.07),
            locate ( 4.07, 2.9, P),

green(line( 4, 1.5, 4, 1)),
            locate ( 3.9, 1, Q),

green(line( 4, 1.5, 4, 4)),
 locate ( 3.9, 4.4, R)
)}}}
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Figure 1. &nbsp;To the Theorem 1

</TD>
 </TR>
</TABLE>Besides of this, &nbsp;the segment &nbsp;EP&nbsp; is the part of the mid-line &nbsp;EF&nbsp; and therefore is parallel to the bases of the trapezoid &nbsp;ABCD.

Thus the segment &nbsp;EP&nbsp; is the line segment drawn from the midpoint &nbsp;E&nbsp; of the lateral side &nbsp;AD&nbsp; of the trapezoid &nbsp;AQRD&nbsp; and parallel to the bases &nbsp;AQ&nbsp; and &nbsp;DR&nbsp; of this trapezoid.

Hence, &nbsp;the segment &nbsp;EP&nbsp; intersects the other lateral side &nbsp;QR&nbsp; of the trapezoid &nbsp;AQRD&nbsp; in its midpoint, &nbsp;in accordance with the Theorem 2 of the lesson <A HREF=http://www.algebra.com/algebra/homework/Polygons/Trapesoids-and-their-mid-lines.lesson>Trapezoids and their mid-lines</A> under the current topic in this site. &nbsp;In other words, &nbsp;the point &nbsp;P&nbsp; is the midpoint of the segment &nbsp;QR, &nbsp;and the segments &nbsp;PQ&nbsp; and &nbsp;PR&nbsp; have the same length.

This is what has to be proved.

<H3>Theorem 2</H3>If a point in a plane is equidistant from the two parallel lines containing the bases of a trapezoid, &nbsp;then the point belongs to the straight line containing the trapezoid's mid-line.

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 <TD>
Proof

Let &nbsp;ABCD&nbsp; be a trapezoid with the bases &nbsp;AB&nbsp; and &nbsp;DC&nbsp; and the mid-line &nbsp;EF&nbsp; (Figure 2).

Let &nbsp;P&nbsp; be a point in a plane equidistant from the straight lines &nbsp;AB&nbsp; and &nbsp;DC.

We need to prove that the point &nbsp;P&nbsp; lies in the straight line containing the mid-line &nbsp;EF

of the trapezoid &nbsp;ABCD.

If the point &nbsp;P&nbsp; coincides with the point &nbsp;E&nbsp; or &nbsp;F&nbsp; then the &nbsp;Theorem&nbsp; statement is true.

Otherwise let us draw the perpendiculars &nbsp;PQ&nbsp; and &nbsp;PR&nbsp; from the point &nbsp;P&nbsp; to the straight

lines &nbsp;AB&nbsp; and &nbsp;DC&nbsp; respectively, &nbsp;where &nbsp;Q&nbsp; and &nbsp;R&nbsp; are the intersection points of the

perpendiculars with these straight lines.

</TD>
 <TD>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{drawing( 280, 160, 0.5, 7.5, 0.5, 4.5, 
            line( 1, 1, 7, 1), 
            line( 1, 1, 3, 4),
            line( 3, 4, 6, 4), 
            line( 7, 1, 6, 4),

locate ( 0.9, 1,   A),
            locate ( 6.9, 1,   B),
            locate ( 6.1, 4.3, C),
            locate ( 2.7, 4.3, D),

line( 2.0, 2.5, 6.5, 2.5),
            locate ( 1.7, 2.7, E),
            locate ( 6.6, 2.7, F),

circle ( 4, 2.5, 0.07, 0.07),
            locate ( 4.07, 2.9, P),

green(line( 4, 1.5, 4, 1)),
            locate ( 3.9, 1, Q),

</TD>
 </TR>
</TABLE>

The perpendiculars &nbsp;PQ&nbsp; and &nbsp;PR&nbsp; lie in one straight line &nbsp;QR.

Consider the trapezoid &nbsp;AQRD&nbsp; with the lateral side &nbsp;QR&nbsp; and the point &nbsp;P&nbsp; in this side. &nbsp;Let us draw the straight segment &nbsp;EP&nbsp; connecting the points &nbsp;E&nbsp; and &nbsp;P.

Since the point &nbsp;P&nbsp; is equidistant from the straight lines &nbsp;AB&nbsp; and &nbsp;DC, &nbsp;the perpendiculars &nbsp;PQ&nbsp; and &nbsp;PR&nbsp; have the same length &nbsp;(see the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Length-and-distance/The-distance-from-a-point-to-a-straight-line.lesson>The distance from a point to a straight line in a plane</A>&nbsp; under the topic &nbsp;Length, distance, coordinates, metric length&nbsp; of the section &nbsp;Geometry&nbsp; in this site).

It means that the point &nbsp;P&nbsp; is the midpoint of the lateral side &nbsp;QR&nbsp; of the trapezoid &nbsp;AQRD.

Hence, the segment &nbsp;EP&nbsp; is the mid-line of the trapezoid &nbsp;AQRD.

It implies that the segment &nbsp;EP&nbsp; is parallel to the bases &nbsp;AQ&nbsp; and &nbsp;DR&nbsp; in accordance with the property of a mid-line of a trapezoid &nbsp;(see the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Polygons/Trapesoids-and-their-mid-lines.lesson>Trapezoids and their mid-lines</A>&nbsp; under the current topic in this site).

Thus we have two straight lines &nbsp;EF&nbsp; and &nbsp;EP&nbsp; passing through the point &nbsp;E&nbsp; and parallel to the bases &nbsp;AB&nbsp; and &nbsp;DC&nbsp; of the trapezoid &nbsp;ABCD. &nbsp;Hence, the straight line &nbsp;EF&nbsp; and &nbsp;EP&nbsp; coincide, and the point &nbsp;P&nbsp; lies on the straight line &nbsp;EF &nbsp;containing the mid-line of the trapezoid &nbsp;ABCD.

The &nbsp;Theorem 2&nbsp; is proved.

<H3>Summary</H3>1. &nbsp;Any point at the mid-line of a trapezoid is equidistant from the two parallel lines containing the bases of the trapezoid.

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

Taken together, &nbsp;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, &nbsp;the problem below shows how the proved properties work.

<H3>Problem 1</H3>In a trapezoid, &nbsp;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. &nbsp;Prove.

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 <TR>
 <TD>
Proof

Let &nbsp;ABCD&nbsp; be a trapezoid with the bases &nbsp;AB&nbsp; and &nbsp;DC&nbsp; and the mid-line &nbsp;EF&nbsp; (Figure 3).&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

Let &nbsp;IJ&nbsp; be an arbitrary straight segment that have one endpoint &nbsp;I&nbsp; on the straight

line containing the base &nbsp;AB&nbsp; of the trapezoid and the other endpoint &nbsp;J&nbsp; on the straight

line containing the other base &nbsp;DC. &nbsp;We need to prove that the mid-line &nbsp;EF&nbsp; of the

trapezoid bisects the segment &nbsp;IJ at the intersection point &nbsp;G.

First of all, &nbsp;the mid-line &nbsp;EF&nbsp; really does intersect the segment &nbsp;IJ, &nbsp;since its endpoints

I&nbsp; and &nbsp;J&nbsp; lie in two different half-planes relative to the mid-line &nbsp;EF.

Keeping this in mind, let us draw the perpendicular segments &nbsp;GK&nbsp; and &nbsp;GL&nbsp; from the

intersection point &nbsp;G&nbsp; to the bases &nbsp;AB&nbsp; and &nbsp;DC&nbsp; correspondingly,

</TD>
 <TD>

{{{drawing( 320, 160, 0.5, 8.5, 0.5, 4.5, 
            line( 1, 1, 8, 1), 
            line( 1, 1, 3, 4),
            line( 3, 4, 7, 4), 
            line( 8, 1, 7, 4),

locate ( 0.9, 1,   A),
            locate ( 7.9, 1,   B),
            locate ( 7.1, 4.3, C),
            locate ( 2.7, 4.3, D),

green(line( 6, 4, 2, 1)),

line( 2, 2.5, 7.5, 2.5),
            locate (1.65, 2.7, E), 
            locate (7.60, 2.7, F),

circle (4.0, 2.5, 0.06, 0.06),
            locate (4.05, 2.45, G),

blue(line( 4, 1, 4, 4)),
            locate (3.9, 4.40, L),
            locate (3.9, 1.00, K),

locate (5.9, 4.40, J),
            locate (1.9, 1.00, I),

line (3.9, 1.6, 4.1, 1.6),
            line (3.9, 1.7, 4.1, 1.7),

line (3.9, 3.2, 4.1, 3.2),
            line (3.9, 3.3, 4.1, 3.3),

arc (4.0, 2.5, 0.8, 0.8, 90, 140),
            arc (4.0, 2.5, 1.0, 1.0, 90, 140),

arc (4.0, 2.5, 0.8, 0.8, 270, 320),
 arc (4.0, 2.5, 1.0, 1.0, 270, 320)
)}}}
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Figure 3. &nbsp;To the &nbsp;Problem 1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
 </TR>
</TABLE>

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

These triangles are right-angled triangles.

They have the congruent legs &nbsp;GK&nbsp; and &nbsp;GL, &nbsp;since the point &nbsp;G&nbsp; belongs to the trapezod's mid-line and, &nbsp;therefore, &nbsp;is equidistant from the straight lines &nbsp;AB&nbsp; and &nbsp;DC.

The triangles &nbsp;GKI&nbsp; and &nbsp;GLJ&nbsp; have congruent acute angles &nbsp;IGK&nbsp; and &nbsp;JGL. &nbsp; These angles are congruent as the vertical angles &nbsp;(see the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Angles/Vertical-Angles-.lesson>Vertical angles</A>&nbsp; under the topic &nbsp;Angles, complementary, supplementary angles&nbsp; of the section &nbsp;Geometry&nbsp; in this site).

Hence, the triangles &nbsp;GKI&nbsp; and &nbsp;GLJ&nbsp; are congruent in accordance with the &nbsp;ASA-test for the triangles congruency &nbsp;(see the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Triangles/Congruence-tests-for-triangles.lesson>Congruence tests for triangles</A>&nbsp; under the topic &nbsp;Triangles&nbsp; of the section &nbsp;Geometry&nbsp; in this site). &nbsp;Therefore, &nbsp;the segments &nbsp;IG&nbsp; and &nbsp;GJ&nbsp; are congruent as the corresponding sides &nbsp;(hypotenuses)&nbsp; in these triangles.

This is what had to be proved.

My other lessons on trapezoids in this site are

&nbsp;&nbsp;&nbsp;&nbsp;-<A HREF =http://www.algebra.com/algebra/homework/Polygons/Trapezoids-and-their-base-angles.lesson>Trapezoids and their base angles</A>,

&nbsp;&nbsp;&nbsp;&nbsp;-<A HREF =http://www.algebra.com/algebra/homework/Polygons/Trapesoids-and-their-mid-lines.lesson>Trapezoids and their mid-lines</A>,

&nbsp;&nbsp;&nbsp;&nbsp;-<A HREF =http://www.algebra.com/algebra/homework/Polygons/Diagonals-of-an-isosceles-trapezoid-are-congruent.lesson>Diagonals of an isosceles trapezoid are congruent</A>,

&nbsp;&nbsp;&nbsp;&nbsp;-<A HREF =http://www.algebra.com/algebra/homework/Polygons/Solving-problems-on-trapezoids.lesson>Solving problems on trapezoids</A>,

&nbsp;&nbsp;&nbsp;&nbsp;-<A HREF =http://www.algebra.com/algebra/homework/Polygons/Solving-problems-on-isoscales-trapezoids.lesson>Solving problems on isosceles trapezoids</A>,

&nbsp;&nbsp;&nbsp;&nbsp;-<A HREF =http://www.algebra.com/algebra/homework/Polygons/Trapezoid-is-uniquely-defined-by-its-sides.lesson>Trapezoid is uniquely defined by the lengths of its sides</A>,

&nbsp;&nbsp;&nbsp;&nbsp;-<A HREF =http://www.algebra.com/algebra/homework/Polygons/HOW-TO-construct-a-trapezoid-given-by-the-lengths-of-its-sides.lesson>HOW TO construct a trapezoid given by the lengths of its sides</A> &nbsp;&nbsp;and

&nbsp;&nbsp;&nbsp;&nbsp;-<A HREF =http://www.algebra.com/algebra/homework/Polygons/PROPERTIES-OF-TRAPEZOIDS.lesson>PROPERTIES OF TRAPEZOIDS</A>

under the current topic, &nbsp;&nbsp;and

&nbsp;&nbsp;&nbsp;&nbsp;-<A HREF =http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-trapezoids.lesson>Solved problems on trapezoids</A> &nbsp;&nbsp;and

&nbsp;&nbsp;&nbsp;&nbsp;-<A HREF =http://www.algebra.com/algebra/homework/word/geometry/Solving-problems-on-isosceles-trapezoids.lesson>Solved problems on isosceles trapezoids</A>

under the topic &nbsp;Geometry&nbsp; of the section &nbsp;Word problems.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/GEOMETRY-your-online-textbook.lesson>GEOMETRY - YOUR ONLINE TEXTBOOK</A>.