Lesson On what segments the angle bisector divides the side of a triangle

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<H2>On what segments the angle bisector divides the side of a triangle</H2>

Theorem
In a triangle, the angle bisector divides the side to which it is drawn, in two segments proportional to the ratio of two other sides of a triangle. Prove.

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Proof

We have a triangle ABC (Figure 1) with the vertexes&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; 
A, B and C. The sides of the triangle BC and AC are of
the length a and b respectively, as shown in the Figure 1. 
The angle bisector CD of the angle C divides the side AB 
into segments AD and BD of the length y and x respectively. 
The Theorem states that {{{x/y = a/b}}}.

We will use the Law of sines. It was established in the 
lessons <A HREF=http://www.algebra.com/algebra/homework/Triangles/Law-of-sines.lesson>Law of sines</A> and <A HREF=http://www.algebra.com/algebra/homework/Triangles/Law-of-sines-the-Geometric_proof.lesson>Law of sines - the Geometric 
Proof</A> that are under the topic Triangles of the section 
Geometry in this site.
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{{{drawing( 300, 200, 0, 6, 0, 4, 
 line( 0.3, 0.5, 5.7, 0.5), 
 line( 0.3, 0.5, 4.0, 3.5),
 line( 4.0, 3.5, 5.7, 0.5),

locate(0.3, 0.5, A),
            locate(5.7, 0.5, B),
            locate(4.0, 3.9, C),

locate(4.9, 2.3, a),
            locate(2.0, 2.3, b),

line (4.0, 3.5, 3.43, 0.5),

locate(3.35, 0.5, D),

locate(1.6, 0.5, y),
 locate(4.5, 0.5, x)
)}}}
&nbsp;&nbsp;&nbsp;&nbsp;Figure 1. To the Theorem 1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
 </TD>
 <TD>
{{{drawing( 300, 200, 0, 6, 0, 4, 
 line( 0.3, 0.5, 5.7, 0.5), 
 line( 0.3, 0.5, 4.0, 3.5),
 line( 4.0, 3.5, 5.7, 0.5),

locate(0.3, 0.5, A),
            locate(5.7, 0.5, B),
            locate(4.0, 3.9, C),

locate(4.9, 2.3, a),
            locate(2.5, 2.7, b),

line (4.0, 3.5, 3.43, 0.5),

locate(3.35, 0.5, D),

locate(1.6, 0.5, y),
            locate(4.5, 0.5, x),

arc(0.3, 0.5, 1.0, 1.0, 318, 360),

arc(5.7, 0.5, 0.9, 0.9, 180, 235),
            arc(5.7, 0.5, 1.1, 1.1, 180, 235),

green(line (2.1,  1.95,  3.43, 0.5)),
            locate(1.9, 2.3, E),

green(line (5.08, 1.55, 3.43, 0.5)),
            locate(5.2, 1.8, F),

locate(2.7, 1.6, z),
 locate(4.2, 1.4, z)
)}}}
&nbsp;&nbsp;Figure 2. To the proof of the Theorem
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</TABLE>

So, let us draw perpendiculars DE and DF from the point D to the sides AC and BC of the triangle till the intersection points E and F respectively (Figure 2). 
Since CD is the angle bisector, the point D is equidistant from the sides AC and BC of the angle ACB (see the lesson <A HREF=http://www.algebra.com/algebra/homework/Triangles/An-angle-bisector-properties.lesson>An angle bisector properties</A> under the topic Triangles of the section Geometry in this site). Therefore, the perpendiculars DE and DF are of equal length. If we denote this length as z, then (Figure 2)

{{{sin(A)=z/y}}}, {{{sin(B)=z/x}}}.

Now apply the Law of sines for the triangle ABC in the form

{{{a/sin(A) = b/sin(B)}}}

and substitute the expressions for {{{sin(A)}}} and {{{sin(B)}}} from the above. After cancellation and transformations you will get

{{{x/y = a/b}}},

which is exactly what have to be demonstrated. The proof is completed.

Below are examples that show how to apply the proved Theorem.

Problem 1

The triangle has the sides of 4 cm, 5 cm and 6 cm long.
Find the length of the segments the angle bisector cuts the side of 5 cm.

Solution

Let {{{x}}} and {{{y}}} be the length of the segments the angle bisector cuts the side of 5 cm.
Then you have the system of two equations for {{{x}}} and {{{y}}}:

{{{system(x + y = 5,
x/y = 4/6)}}}

The latest equation is based on the Theorem above.

To solve the system, express {{{x}}} from the second equation

{{{x = 4/6*y = 2/3*y}}}

and then substitute it to the first equation. You get

{{{2/3*y + y = 5}}},
{{{2y + 3y = 5*3}}},
{{{5y = 15}}},
{{{y=3}}}

and then

{{{x = 2/3*y = 2/3*3 = 2}}}.

Answer. The segments the angle bisector cuts the side of 5 cm are 2 cm and 3 cm long.

Problem 2

The triangle has the perimeter of 60 cm. (60 = 4*(4 + 5 +6) = 16 + 20 + 24)
The angle bisector cuts one side of the triangle into segments of 8 and 12 cm.
Find two other sides of the triangle.

Solution

Let {{{x}}} and {{{y}}} be the length of two other sides of the triangle.
The length of the first side of the triangle is 8 cm + 12 cm = 20 cm.
Hence, the summary length of two other sides of the triangle is equal to 60 cm - 20 cm = 40 cm.
Therefore, you have the system of two equations {{{x}}} and {{{y}}}:

{{{system(x + y = 40,
x/y = 8/12)}}}

The latest equation is based on the Theorem above.

To solve the system, express {{{x}}} from the second equation

{{{x = 8/12*y = 2/3*y}}}

and then substitute it to the first equation. You get

{{{2/3*y + y = 40}}},
{{{2y + 3y = 40*3}}},
{{{5y = 120}}},
{{{y=24}}}

and then

{{{x = 2/3*y = 2/3*24 = 16}}}.

Answer. Two other sides of the triangle are 16 cm and 24 cm long.

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Appendix

Having the Theorem proved, we can easily deduce the formulas&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
expressing the length of segments to which the angle bisector 
cuts the side of the triangle.

Indeed, let {{{x}}} and {{{y}}} be the length of the segments the angle 
bisector drawn between the sides of triangle {{{a}}} and {{{b}}} cuts the side {{{c}}}.
Then you have the system of two equations for {{{x}}} and {{{y}}}:

{{{system(x + y = c,
x/y = a/b)}}}
 </TD>
 <TD>
{{{drawing( 300, 200, 0, 6, 0, 4, 
 line( 0.3, 0.5, 5.7, 0.5), 
 line( 0.3, 0.5, 4.0, 3.5),
 line( 4.0, 3.5, 5.7, 0.5),

locate(0.3, 0.5, A),
            locate(5.7, 0.5, B),
            locate(4.0, 3.9, C),

locate(4.9, 2.3, a),
            locate(2.0, 2.3, b),
            locate(2.7, 0.8, c),

line (4.0, 3.5, 3.43, 0.5),

locate(3.35, 0.5, D),

locate(1.6, 0.5, y),
 locate(4.5, 0.5, x)
)}}}
Figure 1. To the Appendix
 </TD>
 </TR>
</TABLE>

The latest equation is based on the Theorem above.

To solve the system, express {{{x}}} from the second equation

{{{x = a/b*y}}}

and then substitute it to the first equation. We get

{{{a/b*y + y = c}}},

{{{ay + by = b*c}}}   &nbsp;&nbsp;&nbsp;&nbsp;(after multiplication both sides of the previous equation by {{{b}}}),

{{{(a + b)*y = b*c}}},

{{{y=b*c/(a + b)}}}.

Now substitute this expression for {{{y}}} to the formula above for {{{x}}}. We get

{{{x = (a/b)*((b*c)/(a + b)) = (a*c)/(a + b)}}}.

Thus the final result is: The angle bisector drawn between sides {{{a}}} and {{{b}}} of a triangle cuts the side {{{c}}} into segments of the length of {{{x = (a*c)/(a + b)}}} and {{{y = (b*c)/(a + b)}}}.

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>.