Lesson Law of sines - the Geometric Proof

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<H2>Law of Sines - the Geometric Proof</H2>

The Law of Sines is the theorem stating that for any triangle with the angles {{{alpha}}}, {{{beta}}} and {{{gamma}}} and the opposite sides a, b and c the equality takes place

{{{a/sin(alpha)}}} = {{{b/sin(beta)}}} = {{{c/sin(gamma)}}}.

From the one side, this theorem is the Trigonometry theorem.
From the other side, it is the Geometry theorem. 
The trigonometric proof is presented in the lesson <A HREF=http://www.algebra.com/algebra/homework/Triangles/Law-of-sines.lesson>Law of sines</A> under the current topic Triangles of the section Geometry in this site.
In this lesson you will learn the geometric proof of the theorem. 
Moreover, you will learn that these ratios are equal to the diameter of the circumscribed circle for the given triangle.

<TABLE> 
 <TR>
 <TD>
Theorem (the Law of Sines)

In any triangle, the ratio of the length of each side to the sine&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
of the angle opposite that side is the same for all three sides:

{{{a/sin(alpha)}}} = {{{b/sin(beta)}}} = {{{c/sin(gamma)}}} = 2R,

and it is equal to the diameter (or to the doubled radius R) 
of the circumscribed circle for the given triangle.

</TD>
 <TD>
{{{drawing( 250, 200, 0.5, 5.5, 0.5, 4.5, 
 line( 1, 1, 5, 1 ), 
 line( 1, 1, 4, 4 ),
 line( 5, 1, 4, 4), 
 locate ( 1, 1, A),
 locate ( 4.2, 4, B),
 locate ( 5, 1, C),

locate ( 4.7, 2.8, a),
            locate ( 2.0, 2.8, c),
            locate ( 3.0, 1,   b),

locate ( 1.5, 1.4, alpha),
 locate ( 3.8, 3.6, beta),
 locate ( 4.5, 1.4, gamma)
)}}}
Figure 1a. Acute Triangle&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
 </TD>
 <TD>
{{{drawing( 350, 200, 0.5, 7.5, 0.5, 4.5, 
 line( 1, 1, 5, 1 ), 
 line( 1, 1, 7, 4 ),
 line( 5, 1, 7, 4), 
 locate ( 1, 1, A),
 locate ( 7.2, 4, B),
 locate ( 5, 1, C),

locate ( 6.3, 2.7, a),
            locate ( 3.7, 2.8, c),
            locate ( 3.0, 1,   b),

locate ( 1.8, 1.4, alpha),
 locate ( 6.3, 3.6, beta),
 locate ( 4.7, 1.4, gamma)
)}}}
Figure 1b. Obtuse Triangle
 </TD>
 </TR>
</TABLE>Figures 1a and 1b illustrate the Theorem showing an acute and an obtuse triangles.
The side a is opposite to the vertex A and the angle {{{alpha}}}. The side b is opposite to the vertex B and the angle {{{beta}}}. The side c is opposite to the vertex C and the angle {{{gamma}}}.

As it stated, the Theorem is valid for any triangle.

<TABLE> 
 <TR>
 <TD>
The Proof of the Theorem

In Figure 2a, in addition to Figure 1a, are shown the center of the &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
circumscribed circle - the point P, and the three segments PA, PB and 
PC connecting the center with the triangle vertices. &nbsp;&nbsp;The segment PD 
is the continuation of the segment PB and lies in one straight line with PB.

The triangle ACP is isosceles as the segments PA and PC are the equal 
radiuses of the circle. Therefore, the angles ACP and CAP are 
congruent; they are marked by the common symbol {{{theta}}} in Figure 2a. 
Similarly, the triangle ABP is isosceles and, therefore, the angles ABP 
and BAP are congruent; they are marked by the common symbol {{{phi}}}. 
The triangle BCP is isosceles and the angles BCP and CBP are 
congruent; they are marked by the common symbol {{{delta}}}. 
 </TD>
 <TD>
{{{drawing( 250, 250, 0.5, 5.5, -0.5, 4.5, 
 line( 1, 1, 5, 1 ), 
 line( 1, 1, 4, 4 ),
 line( 5, 1, 4, 4), 
 locate ( 0.9, 1, A),
 locate ( 4.0, 4.3, B),
 locate ( 5.0, 1, C),

locate ( 4.7, 2.8, a),
            locate ( 2.0, 2.8, c),
            locate ( 3.0, 1,   b),

locate ( 1.4, 1.4, alpha),
            locate ( 3.8, 3.7, beta),
            locate ( 4.5, 1.4, gamma),

red(line(1,    1, 3, 2)),
        red(line(5,    1, 3, 2)),
        red(line(4.02, 4, 3, 2)),
        red(line(2.72, 1.4,  3.02, 2)),
            locate ( 2.8, 1.5, D),

circle(3.0, 2.0, 0.05, 0.05),
            locate ( 3.15, 2.25, P),

circle(3.0, 2.0, 2.236, 2.236),

arc(1, 1, 2.0, 2.0, 314, 332),
            locate ( 1.9, 1.9, phi),

arc(4, 4, 2.0, 2.0, 114, 134),
            locate ( 3.2, 3.2, phi),

arc(1, 1, 1.9, 1.9, 333, 360),
            arc(1, 1, 2.1, 2.1, 333, 360),
            locate ( 2.05, 1.45, theta),

arc(5, 1, 1.9, 1.9, 180, 207),
            arc(5, 1, 2.1, 2.1, 180, 207),
            locate ( 3.8, 1.45, theta),

arc(5, 1, 1.8,  1.8,  208, 250),
            arc(5, 1, 1.95, 1.95, 208, 250),
            arc(5, 1, 2.1,  2.1,  208, 250),
            locate ( 4.1, 2.0, delta),

arc(4, 4, 1.8, 1.8, 71, 114),
 arc(4, 4, 1.95, 1.95, 71, 114),
 arc(4, 4, 2.1, 2.1, 71, 114),
 locate ( 3.9, 2.9, delta)
)}}}
Figure 2a. To the proof of the Theorem&nbsp;&nbsp;&nbsp;&nbsp; 
 </TD>
 <TD>
{{{drawing( 250, 250, 0.5, 5.5, -0.5, 4.5, 
 line( 1, 1, 5, 1 ), 
 line( 1, 1, 4, 4 ),
 line( 5, 1, 4, 4), 
 locate ( 0.9, 1, A),
 locate ( 4.0, 4.3, B),
 locate ( 5.0, 1, C),

locate ( 4.7, 2.8, a),
            locate ( 2.0, 2.8, c),
            locate ( 3.5, 1,   b/2),
            locate ( 2.3, 1,   b/2),

locate ( 1.4, 1.4, alpha),
            locate ( 3.9, 3.7, beta),
            locate ( 4.5, 1.4, gamma),

red(line(1,    1, 3, 2)),
        red(line(5,    1, 3, 2)),
        red(line(4.02, 4, 3, 2)),
      green(line(3.02, 1, 3.02, 2)),

circle(3.0, 2.0, 0.05, 0.05),
            locate ( 3.15, 2.25, P),

circle(3.0, 2.0, 2.236, 2.236),

locate ( 3.8, 1.9, R),
            locate ( 2.1, 1.9, R),
            locate ( 3.5, 2.9, R),

arc(3, 2, 0.6, 0.6, 90, 150),
            locate ( 2.6, 1.7, beta),

arc(3, 2, 0.6, 0.6, 30, 90),
            locate ( 3.2, 1.7, beta),

locate ( 3.0, 1, E)
)}}} 
Figure 2b. To the proof of the Theorem
 </TD>
 </TR>
</TABLE>

Now, the angle APD is equal to the sum of the angles ABP and PAB as the exterior angle of the triangle ABP. Therefore, the angle APD is equal to the doubled angle {{{phi}}}, or {{{2*phi}}}.
The angle CPD is equal to the sum of the angles BCP and CBP as the exterior angle of the triangle BCP and, therefore, the angle CPD is equal to the doubled angle {{{delta}}}, or {{{2*delta}}}.
The angle APC is equal to the sum of the angles APD and CPD and, therefore, is equal to {{{2*(phi + delta)}}}.
From the other side, the angle ABC is equal to the angle {{{phi + delta}}}, as you can see from the Figure 2a.
Thus, we have found out that the angle APC is the doubled angle ABC.

The rest of the proof is straightforward. From now we will use the Figure 2b. Let us draw the perpendicular PE from the point P to the side AC of the triangle ABC. 
Since the triangle ACP is isosceles, the segment PE, which is the altitude in this triangle, is the median and the angle bisector to the angle APC at the same time. 
This means that the triangle AEP is the right triangle, its leg AE has the length b/2 and the angle APE is half of the angle AEP. 
Taking into account what we proved in the previous paragraph, the later means that the angle APE is congruent to the angle ABC. In other words, the angle APE is equal to {{{beta}}}.

From the triangle AEP, the sines of the angle APE is equal to the ratio of the leg AE length to the hypotenuse AP length, that is sin(LAPE) = {{{b/(2*R)}}}.

Since the angle APE is equal to {{{beta}}}, we can rewrite the last equality as {{{b/sin(beta)}}} = 2R.

Similar arguments work for the two other pairs "the side - the angle" (a, {{{alpha}}}), (c, {{{gamma}}}) of the triangle ABC. This proves the Theorem statement in full.

This theorem is very useful when you calculate the triangle elements. The examples of applications of the Law of sines are presented in the lesson <A HREF=http://www.algebra.com/algebra/homework/Trigonometry-basics/Solve-triangles-using-Law-of-Sines.lesson>Solve triangles using Law of Sines</A> under the topic Trigonometry of the section Algebra-II in this site.

The Law of sines was used in the lesson <A HREF=http://www.algebra.com/algebra/homework/word/geometry/On-what-segments-the-angle-bisector-divides-the-side-of-a-triangle.lesson>On what segments the angle bisector divides the side of a triangle</A> under the topic Geometry of the section Word problems in this site to determine on what segments the angle bisector divides the side of a triangle.

For navigation over the lessons on Properties of Triangles use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/Compendium-of-properties-of-triangles.lesson>Properties of Trianles</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>.