Lesson Solving problems on trapezoids

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<H2>Solving problems on trapezoids</H2>

In this lesson you will find solutions of some typical problems on trapezoids.

<BLOCKQUOTE>Reminder &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).

Trapezoid &nbsp;is a quadrilateral which has two opposite sides parallel and the other two sides non-parallel.

The parallel sides of a trapezoid are called its &nbsp;bases. 
The non-parallel sides of a trapezoid are called its &nbsp;lateral sides&nbsp; or &nbsp;legs.

The angles at the ends of the larger base of a trapezoid are called the &nbsp;base angles .

A &nbsp;mid-line&nbsp; of a trapezoid is the line segment connecting the midpoints of the lateral sides of a trapezoid.</BLOCKQUOTE>

<H3>Problem 1</H3>In a trapezoid, &nbsp;the sum of the interior angles at the ends of a lateral side is equal to &nbsp;180°. &nbsp;Prove.

<TABLE> 
 <TR>
 <TD>
Proof

Let &nbsp;ABCD&nbsp; be a trapezoid with the lateral sides &nbsp;AD&nbsp; and &nbsp;BC &nbsp;(Figure 1).&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

We need to prove that the sum of the interior angles &nbsp;LA&nbsp; and &nbsp;LD&nbsp; at the ends

of the lateral side &nbsp;AD&nbsp; is equal to 180°.

The angles &nbsp;LA&nbsp; and &nbsp;LD&nbsp; are the &nbsp;same side internal angles&nbsp; at the parallel lines

AB&nbsp; and &nbsp;DC&nbsp; (that are the bases of the trapezoid &nbsp;ABCD)&nbsp; and the transverse &nbsp;AD.

It is well known fact that the &nbsp;same side internal angles&nbsp; at the parallel lines

and the transverse are supplementary angles and make &nbsp;180°&nbsp; in sum.

</TD>
 <TD>

{{{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),

arc ( 1, 1, 1.2, 1.2, 300, 360),

arc ( 3, 4, 0.8, 0.8, 0, 120),
 arc ( 3, 4, 1.0, 1.0, 0, 120)
)}}}
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Figure 1. &nbsp;To the &nbsp;Problem 1

</TD>
 </TR>
</TABLE>See the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Angles/Parallel-lines.lesson>Parallel lines</A>&nbsp; under the topic &nbsp;Angles, complementary,

supplementary angles&nbsp; of the section &nbsp;Geometry&nbsp; in this site.

So, &nbsp;the sum of the angles &nbsp;LA&nbsp; and &nbsp;LD&nbsp; is equal to &nbsp;180°. &nbsp;The proof is completed.

Note that similar statement is true for the sum of the interior angles &nbsp;LB&nbsp; and &nbsp;LC&nbsp; at the ends of the other lateral side &nbsp;BC, &nbsp;too.

<H3>Problem 2</H3>In a trapezoid, &nbsp;the bisectors of the interior angles at the ends of a lateral side intersect at the right angle. &nbsp;Prove.
(In other words, the lateral side of the trapezoid is seen at the right angle from the intersection point of the angle bisectors of the angles at the ends of this side)

<TABLE> 
 <TR>
 <TD>
Proof

Let &nbsp;ABCD&nbsp; be a trapezoid and &nbsp;LA&nbsp; and &nbsp;LD&nbsp; be the interior angles at the ends&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

of the lateral side &nbsp;AD &nbsp;(Figure 2). &nbsp;Let &nbsp;AO and &nbsp;DO&nbsp; be the angle bisectors

to the angles &nbsp;LA&nbsp; and &nbsp;LD&nbsp; respectively with the intersection point &nbsp;O.

We need to prove that the angle &nbsp;LAOD&nbsp; is the right angle.

This is very simple.

When solving the &nbsp;Problem 1&nbsp; above, &nbsp;we proved that &nbsp;LA + &nbsp;LD = 180°.

</TD>
 <TD>

{{{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),

arc ( 1, 1, 1.2, 1.2, 300, 360),

arc ( 3, 4, 0.8, 0.8,   0, 120),
            arc ( 3, 4, 1.0, 1.0,   0, 120),

green(line(3, 4, 4.73, 1)),
      green(line(1, 1, 5,    3)),

locate ( 4.0, 2.60, O),

arc ( 3.86, 2.5, 0.65, 0.65,  150, 240),
            arc ( 3.86, 2.5, 0.80, 0.80,  150, 240),
            arc ( 3.86, 2.5, 0.95, 0.95,  150, 240)

)}}}
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Figure 2. &nbsp;To the &nbsp;Problem 2

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

From the other side, &nbsp;the angle &nbsp;LDAO&nbsp; is half of the angle &nbsp;LA&nbsp; and the angle &nbsp;LODA&nbsp; is half of the angle &nbsp;LD, &nbsp;since &nbsp;AO&nbsp; and &nbsp;DO&nbsp; are the angle bisectors to the

angles &nbsp;LA&nbsp; and &nbsp;LD. &nbsp;Therefore, &nbsp;the sum of the angles &nbsp;LDAO&nbsp; and &nbsp;LODA&nbsp; is half of &nbsp;180°, &nbsp;i.e. &nbsp;90°: &nbsp;LDAO&nbsp; + &nbsp;LODA = 90°. &nbsp;Hence, &nbsp;the measure of the angle

LAOD&nbsp; is equal to &nbsp;180° - (LDAO&nbsp; + &nbsp;LODA) = 180° - 90° = 90°.

The proof is completed.

<H3>Problem 3</H3>In a trapezoid, &nbsp;the bisectors of the interior angles at the ends of a lateral side intersect at the point which lies on the mid-line of the trapezoid. &nbsp;Prove.

<TABLE> 
 <TR>
 <TD>
Proof

Let &nbsp;ABCD&nbsp; be a trapezoid and &nbsp;LA&nbsp; and &nbsp;LD&nbsp; be the interior angles at the&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

ends of the lateral side &nbsp;AD &nbsp;(Figure 3a). &nbsp;Let &nbsp;AO and &nbsp;DO&nbsp; be the angle

bisectors to the angles &nbsp;LA&nbsp; and &nbsp;LD&nbsp; respectively with the intersection

point &nbsp;O.

We need to prove that the intersection point &nbsp;O&nbsp; lies on the mid-line &nbsp;EF&nbsp;

of the trapezoid &nbsp;ABCD.

The angle bisector &nbsp;AO of the angle &nbsp;LA&nbsp; is the locus of points equidistant

from the sides &nbsp;AB&nbsp; and &nbsp;AD &nbsp; &nbsp;(see the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Triangles/An-angle-bisector-properties.lesson>An angle bisector properties</A>&nbsp;

under the topic &nbsp;Triangles&nbsp; of the section &nbsp;Geometry&nbsp; in this site).

</TD>
 <TD>

{{{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),

arc ( 1, 1, 1.2, 1.2, 300, 360),

arc ( 3, 4, 0.8, 0.8,   0, 120),
            arc ( 3, 4, 1.0, 1.0,   0, 120),

green(line( 3, 4, 4.73, 1)),
      green(line( 1, 1, 5,    3)),

locate ( 3.85, 2.90, O),

blue(line( 2, 2.47, 6.5, 2.47)),
 locate (1.65, 2.7, E), 
 locate (6.60, 2.7, F)
)}}}
&nbsp;&nbsp;&nbsp;&nbsp;Figure 3a. &nbsp;To the &nbsp;Problem 3&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
 <TD>

{{{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),

arc ( 1, 1, 1.2, 1.2, 300, 360),

arc ( 3, 4, 0.8, 0.8,   0, 120),
            arc ( 3, 4, 1.0, 1.0,   0, 120),

green(line( 3, 4, 4.73, 1)),
      green(line( 1, 1, 5,    3)),

locate ( 3.85, 2.90, O),

blue(line( 2, 2.47, 6.5, 2.47)),
            locate (1.65, 2.7, E), 
            locate (6.60, 2.7, F),

red(arc (3.84, 2.48, 3.0, 3.0, 45, 315))
)}}}
Figure 3b. To the solution of the Problem 3

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

The angle bisector &nbsp;DO of the angle &nbsp;LD&nbsp; is the locus of points equidistant from the sides &nbsp;AD&nbsp; and &nbsp;DC&nbsp; by the same reason.

Hence, &nbsp;the intersection point &nbsp;O&nbsp; of these two angle bisectors is equidistant from the three segments &nbsp;AB, &nbsp;AD&nbsp; and &nbsp;DC. &nbsp;&nbsp;

(Visually, &nbsp;this means that the point &nbsp;O&nbsp; is the center of the circle which touches the segments &nbsp;AB, &nbsp;AD&nbsp; and &nbsp;DC, &nbsp;see the &nbsp;Figure 3b).

In particular, the intersection point &nbsp;O&nbsp; is equidistant from the segments &nbsp;AB&nbsp; and &nbsp;DC.

It implies that the point &nbsp;O&nbsp; lies on the mid-line of the trapezoid &nbsp;ABCD, &nbsp;since the mid-line of a trapezoid is the locus of point equidistant from the parallel straight lines

containing the bases of the trapezoid &nbsp;(see the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Polygons/Mid-line-of-a-trapezoid-is-the-locus-of-points-equidistant-from-its-bases.lesson>Mid-line of a trapezoid is the locus of points equidistant from its bases</A>&nbsp; under the current topic in this site).

The proof is completed.

<H3>Problem 4</H3>In a trapezoid, &nbsp;the distance between midpoints of its diagonals is half of the difference of the lengths of the larger and the shorter bases. &nbsp;Prove.

<TABLE> 
 <TR>
 <TD>
Proof

Let &nbsp;ABCD&nbsp; be a trapezoid with the bases &nbsp;AB&nbsp; and &nbsp;DC&nbsp; and with the diagonals &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

AC&nbsp; and &nbsp;BD&nbsp; (Figure 4). &nbsp;Let &nbsp;EF&nbsp; be the mid-line of the trapezoid &nbsp;ABCD&nbsp; and

G, &nbsp;H&nbsp; be the intersection points of the mid-line &nbsp;EF&nbsp; with the diagonals

AC&nbsp; and &nbsp;BD&nbsp; respectively.

Since &nbsp;EF&nbsp; is the mid-line, &nbsp;it is parallel to the trapezoid bases and passes through

the midpoint &nbsp;E&nbsp; of the lateral side &nbsp;AD. &nbsp;Hence, the segment &nbsp;EG&nbsp; is parallel to

the base &nbsp;DC&nbsp; and met the diagonal &nbsp;AC&nbsp; at its midpoint &nbsp;G &nbsp;(see the &nbsp;Theorem 3

of the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Triangles/The-line-segment-joining-the-midpoints-of-two-sides-of-a-triangle.lesson>The line segment joining the midpoints of two sides of a triangle</A>&nbsp;

under the topic &nbsp;Triangles&nbsp; of the section &nbsp;Geometry&nbsp; in this site).

</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),

locate ( 4.9, 4.42,   d),
            locate ( 4.4, 1.42,   a),

green(line( 3, 4, 8, 1)),
      green(line( 1, 1, 7, 4)),

blue(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 (4.0, 2.45, G),

circle (5.5, 2.5, 0.06, 0.06),
 locate (5.4, 2.45, H)
)}}}
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Figure 4. &nbsp;To the &nbsp;Problem 4&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

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

Thus the segment &nbsp;EG&nbsp; is the midpoint line of the triangle &nbsp;ADC, &nbsp;and the intersection point &nbsp;G&nbsp; of the segments &nbsp;EF&nbsp; and &nbsp;AC&nbsp; is the midpoint of the diagonal &nbsp;AC.

The length of the segment &nbsp;EG&nbsp; is half of the length of the base &nbsp;DC: &nbsp;&nbsp; |EG| = {{{abs(DC)/2}}} = {{{d/2}}}.

By the similar reasons, the segment &nbsp;HF&nbsp; is the midpoint line of the triangle &nbsp;BDC, &nbsp;and the intersection point &nbsp;H&nbsp; of the segments &nbsp;EF&nbsp; and &nbsp;BD&nbsp; is the midpoint of

the diagonal &nbsp;BD.

The length of the segment &nbsp;HF&nbsp; is half of the length of the base &nbsp;DC: &nbsp;&nbsp; |HF| = {{{abs(DC)/2}}} = {{{d/2}}}.

We need to find the length of the segment &nbsp;GH.

It is easy to do now. &nbsp;The length of the segment &nbsp;GH&nbsp; is equal to the length of the mid-line &nbsp;EF&nbsp; minus the length of the segment &nbsp;EG&nbsp; minus the length of the segment &nbsp;HF:

|GH| = |EF| - |EG| - |HF| = {{{(a + d)/2}}} - {{{d/2}}} - {{{d/2}}} = {{{(a - d)/2}}},

where &nbsp;a&nbsp; is the length of the larger base &nbsp;AB&nbsp; and &nbsp;d&nbsp; is the length of the shorter base &nbsp;DC&nbsp; of the trapezoid &nbsp;ABCD. &nbsp;This is exactly what has 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/Mid-line-of-a-trapezoid-is-the-locus-of-points-equidistant-from-its-bases.lesson>Mid-line of a trapezoid is the locus of points equidistant from its bases</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>.