Lesson One property of a trapezoid

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<H2>One property of a trapezoid</H2>

<H3>Problem 1</H3>In a trapezoid, &nbsp;the straight line passing through the intersection point of the two lateral sides and the intersection point of its diagonals bisects each base of the trapezoid. &nbsp;Prove.

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Proof

Let &nbsp;ABCD&nbsp; be a trapezoid &nbsp;(Figure 1); &nbsp;G&nbsp; be the&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

intersection point of its lateral sides &nbsp;AD&nbsp; and &nbsp;BC;

E&nbsp; and &nbsp;F&nbsp; be the midpoints of its bases &nbsp;AB&nbsp; and &nbsp;DC;

O&nbsp; be the intersection point of its diagonals &nbsp;AC&nbsp; and &nbsp;BD.

The statement is that the four points &nbsp;G, &nbsp;E, &nbsp;O and &nbsp;F

lie in one straight line.

We will prove the statement in two steps.

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&nbsp;&nbsp;&nbsp;&nbsp;{{{drawing( 210, 210, 0.5, 7.5, 0.5, 7.5, 
            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, 4.4, C),
            locate ( 2.7, 4.4, D),

green(line( 1, 1, 6, 4)), 
      green(line( 3, 4, 7, 1)),

green(line( 3, 4, 5, 7)), 
      green(line( 6, 4, 5, 7)),

red(line( 5, 7, 4, 1)),

locate ( 4.9,  7.4,  G),
            locate ( 4.25, 4.55, E),
            locate ( 3.9,  1.0,  F),

locate ( 4.3, 2.9, O)
)}}}
&nbsp;&nbsp;&nbsp;&nbsp;Figure 1. &nbsp;To the &nbsp;Problem 1

</TD>
 <TD>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{drawing( 210, 210, 0.5, 7.5, 0.5, 7.5, 
            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, 4.4, C),
            locate ( 2.7, 4.4, D),

green(line( 3, 4, 5, 7)), 
      green(line( 6, 4, 5, 7)),

red(line( 5, 7, 4, 1)),

locate ( 4.9, 7.4, G),
 locate ( 4.25, 4.55, E),
 locate ( 3.9, 1.0, F)
)}}}
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Figure 2a. &nbsp;To the &nbsp;Step 1

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locate ( 0.9, 1,   A),
            locate ( 6.9, 1,   B),
            locate ( 6.1, 4.4, C),
            locate ( 2.7, 4.4, D),

green(line( 1, 1, 6, 4)), 
      green(line( 3, 4, 7, 1)),

red(line( 4.5, 4, 4, 1)),

locate ( 4.4, 4.55, E),
            locate ( 3.9, 1.0,  F),

locate ( 4.3, 2.9, O)
)}}}
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Figure 2b. &nbsp;To the &nbsp;Step 2

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Step 1&nbsp; (Figure 2a). &nbsp;The three points &nbsp;G, &nbsp;E&nbsp; and &nbsp;F&nbsp; lie in one straight line. &nbsp;In other words, &nbsp;for any trapezoid the intersection point of its lateral sides and the midpoints of its bases lie in one straight line.

It was proved in the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Triangles/One-property-of-a-median-in-a-triangle.lesson>One property of a median in a triangle</A>&nbsp; that the median of a triangle drawn from the vertex to the base bisects every segment parallel to the base and con-

necting two other sides. &nbsp;In a triangle &nbsp;ABG&nbsp; the straight segment &nbsp;GF&nbsp; is the median, &nbsp;while the segment &nbsp;DC&nbsp; is parallel to the base &nbsp;AB. &nbsp;Hence, the median &nbsp;GF&nbsp; intersects the base &nbsp;DC&nbsp; of the trapezoid at the midpoint of the segment &nbsp;DC. &nbsp;Thus the intersection points of the trapezoid's lateral sides and the midpoints of its two bases lie in one straight line. &nbsp;The &nbsp;Step 1&nbsp; statement is proved.

Step 2&nbsp; (Figure 2b). &nbsp;The three points &nbsp;E, &nbsp;O&nbsp; and &nbsp;F&nbsp; lie in one straight line. &nbsp;In other words, &nbsp;for any trapezoid the straight line passing through the midpoint of a base and the intersection point of its diagonals bisects the other base.

Indeed, let the straight line &nbsp;EF&nbsp; (Figure 2b)&nbsp; passes through the midpoint &nbsp;E&nbsp; of the base &nbsp;DC&nbsp; and the intersection point &nbsp;O&nbsp; of the diagonals. &nbsp;Then the triangles &nbsp;OEC&nbsp; and &nbsp;OFA&nbsp; are similar in accordance to the &nbsp;AA&nbsp; similarity test for triangles, &nbsp;because they have two pairs of congruent angles. &nbsp;Therefore, &nbsp;the corresponding sides of these triangles ae proportional:

{{{abs(OF)/abs(OE)}}} = {{{abs(AF)/abs(EC)}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(1)

Also, &nbsp;the triangles &nbsp;OED&nbsp; and &nbsp;OFB&nbsp; are similar by the same reason. &nbsp;Therefore, &nbsp;their corresponding sides are proportional too:

{{{abs(OF)/abs(OE)}}} = {{{abs(FB)/abs(DE)}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(2)

From proportions &nbsp;(1)&nbsp; and &nbsp;(2)&nbsp; we have

{{{abs(AF)/abs(EC)}}} = {{{abs(FB)/abs(DE)}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(3)

Since &nbsp;|EC| = |DE|, &nbsp;the proportion &nbsp;(3)&nbsp; implies &nbsp;|AF| = |FB|. &nbsp;It means that the point &nbsp;F&nbsp; is the midpoint of the segment &nbsp;AB.

Thus the midpoints of the bases of the trapezoid and the intersection point of its diagonals lie in one straight line. &nbsp;The &nbsp;Step 2&nbsp; statement is proved.

From the proved statements of the &nbsp;Step 1&nbsp; and the &nbsp;Step 2&nbsp; we get the proof to the statement of the &nbsp;Problem 1.

<H3>Summary</H3>For any trapezoid, &nbsp;the midpoints of its bases, &nbsp;the intersection point of its diagonals and the intersection point of its lateral sides lie in one straight line.

My other lessons on similar triangles in this site are

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/Similar-triangles.lesson>Similar triangles</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/Similarity-tests-for-triangles.lesson>Similarity tests for triangles</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/Proofs-of-Similarity-tests-for-triangles.lesson>Proofs of Similarity tests for triangles</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/In-a-triangle-a-straight-line-parallel-to-its-side-cuts-off-a-similar-triangle.lesson>In a triangle a straight line parallel to its side cuts off a similar triangle</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/Problems-on-similar-triangles.lesson>Problems on similar triangles</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/Similarity-tests-for-right-angled-triangles.lesson>Similarity tests for right-angled triangles</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/Problems-on-similarity-for-right-angled-triangles.lesson>Problems on similarity for right-angled triangles</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/Problems-on-similarity-for-right-angled-and-acute-triangles.lesson>Problems on similarity for right-angled and acute triangles</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/One-property-of-a-median-in-a-triangle.lesson>One property of a median in a triangle</A> &nbsp;&nbsp;and

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Triangles/Miscellaneous-problems-on-similar-triangles.lesson>Miscellaneous problems on similar triangles</A>

under the current topic, &nbsp;and

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