Lesson Consecutive angles of a parallelogram

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<H2>Consecutive angles of a parallelogram</H2>

Two interior angles of a parallelogram are called the consecutive angles if some side of the parallelogram is the common side of these two angles.

<TABLE>
 <TR>
 <TD>
<BLOCKQUOTE>Figure 1&nbsp; shows the parallelogram &nbsp;ABCD.
The consecutive angles of the parallelogram &nbsp;ABCD&nbsp; are the angles&nbsp;&nbsp;&nbsp;&nbsp; 
LA and LB; &nbsp;LB and LC; &nbsp;LC and LD; &nbsp;LA and LD. 
The angles &nbsp;LA&nbsp; and &nbsp;LC&nbsp; are not the consecutive angles; they are opposite angles. 
The angles &nbsp;LB&nbsp; and &nbsp;LD&nbsp; are not the consecutive angles; they are opposite angles.</BLOCKQUOTE>
 </TD>
 <TD>
{{{drawing( 280, 160, 0.5, 7.5, 0.5, 4.5, 
 line( 1.0, 1.0, 6.0, 1.0), 
 line( 2.0, 4.0, 7.0, 4.0),
 line( 1.0, 1.0, 2.0, 4.0),
 line( 6.0, 1.0, 7.0, 4.0),

locate(1.0, 1.0, A),
 locate(6.0, 1.0, B),
 locate(7.0, 4.4, C),
 locate(2.0, 4.4, D)
)}}}
Figure 1. Explaining consecutive angles of a parallelogram
 </TD>
 </TR>
</TABLE>Theorem 1
The sum of any two consecutive angles of a parallelogram is equal to the straight angle (180�).<TABLE>
 <TR>
 <TD><BLOCKQUOTE>Reminder (see the lesson <A HREF=http://www.algebra.com/algebra/homework/Angles/Angles-basics.lesson>Angles basics</A> under the topic Angles, complementary, supplementary angles of the 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;section Geometry in this site). 
A Straight Angle is the angle formed by the two rays that are lying in same a straight line 
and have opposite directions. 
Each side of a straight angle is a continuation of the other one (Figure 2).

A Straight Angle is 180�, or {{{pi}}} radians.
All straight angles are equal, or congruent.</BLOCKQUOTE>
 </TD>
 <TD>
{{{drawing( 300, 60, 0, 5, 0, 2, 
 line( 0.5, 1, 4.5, 1),

circle ( 1.0, 1, 0.04),
            circle ( 4.0, 1, 0.04),
            circle ( 2.5, 1, 0.04),

locate ( 1.0, 0.9, A),
 locate ( 4.0, 0.9, B),
 locate ( 2.5, 0.9, O)
)}}}
&nbsp;&nbsp;&nbsp;&nbsp;Figure 2. Straight angle AOB. 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The line AOB is a straight line.
 </TD>
 </TR>
</TABLE><TABLE><TR><TD>Proof to the Theorem 1

Let us consider two consecutive angles DAB and ABC first (Figure 3).

Draw the straight line &nbsp;AE&nbsp; as the continuation of the side &nbsp;AB&nbsp; of the&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

parallelogram &nbsp;ABCD. &nbsp;Then the angle &nbsp;CBE&nbsp; is congruent to the

angle &nbsp;DAB&nbsp; as these angles are the corresponding angles at the

parallel lines &nbsp;AD&nbsp; and &nbsp;BC&nbsp; and the transverse &nbsp;AE&nbsp; (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).

The angles &nbsp;ABC&nbsp; and &nbsp;CBE&nbsp; are adjacent supplementary angles 
and make in sum the straight angle &nbsp;ABE&nbsp; of &nbsp;180�&nbsp; (see the lesson

<A HREF=http://www.algebra.com/algebra/homework/Angles/Angles-basics.lesson>Angles basics</A>&nbsp; under the topic &nbsp;Angles, complementary, supplementary

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

</TD>
 <TD>
{{{drawing( 300, 180, 0.5, 8.0, 0.5, 5.0, 
 line( 1.0, 1.0, 6.0, 1.0), 
 line( 2.0, 4.0, 7.0, 4.0),
 line( 1.0, 1.0, 2.0, 4.0),
 line( 6.0, 1.0, 7.0, 4.0),

locate(1.0, 1.0, A),
            locate(6.0, 1.0, B),
            locate(7.0, 4.4, C),
            locate(2.0, 4.4, D),

arc(1.0, 1.0, 1.12, 1.12, 290, 360),

arc(6.0, 1.0, 1.0, 1.0, 180, 287),
            arc(6.0, 1.0, 1.2, 1.2, 180, 287),

green(line( 6.0, 1.0, 7.3, 1.0)), 
 locate(7.2, 1.0, E),
 arc(6.0, 1.0, 1.12, 1.12, 290, 360)
)}}}
Figure 3. To the proof of the Theorem 1&nbsp;&nbsp;&nbsp;&nbsp; 
for the consecutive angles DAB and ABC
 </TD>
 <TD>
{{{drawing( 300, 180, 0.5, 8.0, 0.5, 5.0, 
 line( 1.0, 1.0, 6.0, 1.0), 
 line( 2.0, 4.0, 7.0, 4.0),
 line( 1.0, 1.0, 2.0, 4.0),
 line( 6.0, 1.0, 7.0, 4.0),

locate(1.0, 1.0, A),
            locate(6.0, 1.0, B),
            locate(7.1, 4.2, C),
            locate(2.0, 4.4, D),

arc(6.0, 1.0, 1.0, 1.0, 180, 287),
            arc(6.0, 1.0, 1.2, 1.2, 180, 287),

green(line( 6.0, 1.0, 7.3, 4.9)), 
            locate(7.3, 4.9, F),
            arc(7.0, 4.0, 1.0, 1.0, 180, 287),
            arc(7.0, 4.0, 1.2, 1.2, 180, 287),

arc(7.0, 4.0, 1.1, 1.1, 107, 180)
)}}}
Figure 4. To the proof of the Theorem 1 
for the consecutive angles ABC and DCB
 </TD>
 </TR>
</TABLE>Therefore, &nbsp;two consecutive angles &nbsp;DAB&nbsp; and &nbsp;ABC&nbsp; are non-adjacent supplementary angles and make in sum the straight angle of &nbsp;180�.
So, &nbsp;the &nbsp;Theorem 1&nbsp; is proved for the consecutive angles &nbsp;DAB&nbsp; and &nbsp;ABC.

Now, &nbsp;let us consider two other consecutive angles &nbsp;ABC&nbsp; and &nbsp;BCD&nbsp; (Figure 4).
Draw the straight line &nbsp;BF&nbsp; as the continuation of the side &nbsp;BC&nbsp; of the parallelogram &nbsp;ABCD. &nbsp;Then the angle &nbsp;DCF&nbsp; is congruent to the angle &nbsp;ABC&nbsp; as these angles are the corresponding angles at the parallel lines &nbsp;DC&nbsp; and &nbsp;AB&nbsp; and the transverse &nbsp;BF. &nbsp;The angles &nbsp;BCD&nbsp; and &nbsp;DCF&nbsp; are adjacent supplementary angles and make in sum the straight angle &nbsp;BCF&nbsp; of &nbsp;180�.
Therefore, &nbsp;two consecutive angles &nbsp;ABC&nbsp; and &nbsp;BCD&nbsp; are non-adjacent supplementary angles and make in sum the straight angle of &nbsp;180�.
So, &nbsp;the &nbsp;Theorem 1&nbsp; is proved for the consecutive angles &nbsp;ABC&nbsp; and &nbsp;BCD&nbsp; too.

For the rest of consecutive angles the proof is similar. &nbsp;Thus the &nbsp;Theorem 1&nbsp; is fully proved.

The converse statement to the &nbsp;Theorem 1&nbsp; is valid too.

Theorem 2
If in a quadrilateral the sum of any two consecutive angles is equal to the straight angle &nbsp;(180�), &nbsp;the quadrilateral is a parallelogram .

<TABLE>
<TR>
<TD>
Proof

For the proof we will use the same Figure 3 and Figure 4 from the above.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
For your convenience I simply put the copies of these Figures to the 
appropriate place under the names &nbsp;Figure 5&nbsp; and &nbsp;Figure 6.

Let us consider two consecutive angles &nbsp;DAB&nbsp; and &nbsp;ABC&nbsp; (Figure 5).

The sum of these two angles is equal to &nbsp;180�&nbsp; by the condition.

Draw the straight line &nbsp;AE&nbsp; as the continuation of the side &nbsp;AB&nbsp; of the

parallelogram &nbsp;ABCD. &nbsp;The sum of the angles &nbsp;ABC&nbsp; and &nbsp;CBE&nbsp; is equal

to &nbsp;180�&nbsp; as they are adjacent supplementary angles. &nbsp;It means that 
the angle &nbsp;DAB&nbsp; is congruent to the angle &nbsp;CBE. &nbsp;This implies, &nbsp;in turn,

that the straight lines &nbsp;AD&nbsp; and &nbsp;BC&nbsp; are parallel as the angles &nbsp;DAB

locate(1.0, 1.0, A),
            locate(6.0, 1.0, B),
            locate(7.0, 4.4, C),
            locate(2.0, 4.4, D),

arc(1.0, 1.0, 1.12, 1.12, 290, 360),

arc(6.0, 1.0, 1.0, 1.0, 180, 287),
            arc(6.0, 1.0, 1.2, 1.2, 180, 287),

green(line( 6.0, 1.0, 7.3, 1.0)), 
 locate(7.2, 1.0, E),
 arc(6.0, 1.0, 1.12, 1.12, 290, 360)
)}}}
Figure 5. To the proof of the Theorem 2.&nbsp;&nbsp;&nbsp;&nbsp; 
The consecutive angles DAB and ABC
 </TD>
 <TD>
{{{drawing( 300, 180, 0.5, 8.0, 0.5, 5.0, 
 line( 1.0, 1.0, 6.0, 1.0), 
 line( 2.0, 4.0, 7.0, 4.0),
 line( 1.0, 1.0, 2.0, 4.0),
 line( 6.0, 1.0, 7.0, 4.0),

locate(1.0, 1.0, A),
            locate(6.0, 1.0, B),
            locate(7.1, 4.2, C),
            locate(2.0, 4.4, D),

arc(6.0, 1.0, 1.0, 1.0, 180, 287),
            arc(6.0, 1.0, 1.2, 1.2, 180, 287),

green(line( 6.0, 1.0, 7.3, 4.9)), 
            locate(7.3, 4.9, F),
            arc(7.0, 4.0, 1.0, 1.0, 180, 287),
            arc(7.0, 4.0, 1.2, 1.2, 180, 287),

arc(7.0, 4.0, 1.1, 1.1, 107, 180)
)}}}
Figure 6. To the proof of the Theorem 2.
The consecutive angles ABC and DCB
 </TD>
 </TR>
</TABLE>and &nbsp;CBE&nbsp; are the corresponding angles of these straight lines and the transverse &nbsp;AB&nbsp; (see the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Angles/Parallel-lines.lesson>Parallel lines</A>&nbsp; in this site).

Let us consider now two other consecutive angles &nbsp;ABC&nbsp; and &nbsp;BCD&nbsp; (Figure 6).

The sum of these two angles is equal to &nbsp;180�&nbsp; by the condition.

Draw the straight line &nbsp;BF&nbsp; as the continuation of the side &nbsp;AB&nbsp; of the parallelogram &nbsp;ABCD. &nbsp;The sum of the angles &nbsp;BCD&nbsp; and &nbsp;DBF&nbsp; is equal to &nbsp;180�&nbsp; as they are adjacent supplementary angles. &nbsp;It means that the angle &nbsp;ABC&nbsp; is congruent to the angle &nbsp;DCF. &nbsp;This implies, &nbsp;in turn, &nbsp;that the straight lines &nbsp;AB&nbsp; and &nbsp;DC&nbsp; are parallel as the angles &nbsp;ABC&nbsp; and &nbsp;DCF&nbsp; are the corresponding angles of these straight lines and the transverse &nbsp;BC&nbsp; (see the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Angles/Parallel-lines.lesson>Parallel lines</A>&nbsp; in this site).

Thus, &nbsp;we have proved that under the given conditions the quadrilateral &nbsp;ABCD&nbsp; has the opposite sides parallel. &nbsp;Hence, &nbsp;it is parallelogram. &nbsp;The &nbsp;Theorem 2&nbsp; is proved.

Problem

In a parallelogram, one of four interior angles is equal to &nbsp;56�. 
Find the rest interior angles of a parallelogram.

Solution

Two other consecutive angles of the parallelogram are equal to &nbsp;180� - 56� = 124� according to the &nbsp;Theorem 1.
The opposite angle to the given one of &nbsp;56� is congruent to it according to the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Parallelograms/Opposite-angles-of-a-parallelogram-are-congruent.lesson>Opposite angles of a parallelogram</A>&nbsp; under the current topic &nbsp;Parallelograms&nbsp; of the section Geometry&nbsp; in this site. &nbsp;So, &nbsp;the opposite angle is equal to &nbsp;56�.

Answer. Three other interior angles of the parallelogram are equal to &nbsp;124�, &nbsp;56�&nbsp; and &nbsp;124�.

This example shows that if you know one angle of a parallelogram, &nbsp;you can easily calculate all three other angles of the parallelogram.

My other lessons on parallelograms in this site are

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Parallelograms/In-a-parallelogram-each-diagonal-divides-it-in-two-congruent-triangles.lesson>In a parallelogram, each diagonal divides it in two congruent triangles</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Parallelograms/Properties-of-the-sides-of-a-parallelogram.lesson>Properties of the sides of a parallelogram</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Parallelograms/Properties-of-the-sides-of-parallelograms.lesson>Properties of the sides of parallelograms</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Parallelograms/Properties-of-diagonals-of-parallelograms.lesson>Properties of diagonals of parallelograms</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Parallelograms/Opposite-angles-of-a-parallelogram-are-congruent.lesson>Opposite angles of a parallelogram</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Midpoints-of-a-quadrilateral-are-vertices-of-the-parallelogram.lesson>Midpoints of a quadrilateral are vertices of the parallelogram</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/The-length-of-diagonals-of-a-parallelogram.lesson>The length of diagonals of a parallelogram</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Parallelograms/Remarcable-advanced-problems-on-parallelograms.lesson>Remarcable advanced problems on parallelograms</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/How-to-solve-problems-on-the-parallelogram-sides-measures-Examples.lesson>HOW TO solve problems on the parallelogram sides measures - Examples</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/How-to-solve-problems-on-the-angles-of-parallelograms.lesson>HOW TO solve problems on the angles of parallelograms - Examples</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Parallelograms/PROPERTIES-OF-PARALLELOGRAMS.lesson>PROPERTIES OF PARALLELOGRAMS</A>

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>.