Lesson An angle bisector properties

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<H2>An angle bisector properties</H2>

In this lesson you will learn about the properties of an angle bisector.
I will refer to the lessons <A HREF=http://www.algebra.com/algebra/homework/Triangles/Congruence-tests-for-triangles.lesson>Congruence_tests_for_triangles</A> and <A HREF=http://www.algebra.com/algebra/homework/Triangles/Isosceles-triangles.lesson>Isosceles triangles</A> that are under the current topic Triangles in the section Geometry in this site, 
so I suppose you are familiar with the contents of these lessons.

<BLOCKQUOTE>Reminder
An angle bisector is the ray passing through the angle vertex, which divides (separates, cuts) the angle in the two congruent angles.</BLOCKQUOTE>

Theorem 1 (Point on angle bisector theorem)

If a point is located on an angle bisector, then it is equidistant from the sides of the angle.

<TABLE>
 <TR>
 <TD>
Proof

In the Figure 1 the angle BAC is shown. AB and AC are the angle sides,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
and the point A is the angle vertex. The angle bisector AP is shown too in
the Figure 1 as the red ray. It divides the angle BAC in the two congruent 
angles BAP and PAC. The point P is an arbitrary point on the angle 
bisector. We need to prove that the point P is equidistant from the sides 
AB and AC of the angle.

Let me remind you that the distance from the point P to the straight line 
is the length of the perpendicular from this point to the line. So, let us draw 
the perpendiculars PD and PE from the point P to the angle sides AB and AC. 
We need to prove that the segments PD and PE are of equal length.
 </TD>
 <TD>
{{{drawing( 200, 200, 0, 4, 0, 4, 
 line( 0.3, 0.5, 3.7, 0.5), 
 line( 0.3, 0.5, 2.0, 3.5),
 red(line( 0.3, 0.5, 3.7, 2.5)),

circle(2.5, 1.8, 0.05, 0.05),

locate(0.3, 0.5, A),
            locate(3.7, 0.5, B),
            locate(2.0, 3.9, C),
            locate(2.5, 2.2, P),

green(line (2.5, 1.8, 2.5, 0.5)),
            locate(2.5, 0.5, D),

green(line (2.5, 1.8, 1.4, 2.45)),
            locate(1.2, 2.8, E),

arc(0.3, 0.5, 0.8, 0.8, 300, 330),
            arc(0.3, 0.5, 0.9, 0.9, 300, 330),

arc(0.3, 0.5, 1.0, 1.0, 330, 360),
            arc(0.3, 0.5, 1.1, 1.1, 330, 360),

arc(2.5, 1.8, 0.8, 0.8, 150, 210),

arc(2.5, 1.8, 1.0, 1.0, 90, 150)
)}}}
Figure 1. To the Theorem 1
 </TD>
 </TR>
</TABLE>

Consider the triangles ADP and AEP. These are the right triangles. 
Since AP is the angle bisector, the angles DAP and PAE are congruent. 
It implies that the angles APD and APE are congruent as the non-adjacent complementary angles to the angles DAP and PAE in the right triangles ADP and AEP.

Thus, the triangles ADP and AEP have the common side AP and the congruent angles DAP and PAE, APD and APE that include this side.
Therefore, the triangles ADP and AEP are congruent in accordance to the postulate P2 (ASA) of the lesson <A HREF=http://www.algebra.com/algebra/homework/Triangles/Congruence-tests-for-triangles.lesson>Congruence tests for triangles</A>, which is under 
the current topic Triangles in the section Geometry in this site.
Hence, the segments PD and PE are of equal length as the corresponding sides of these triangles.
The proof is completed.

The opposite statement to the Theorem 1 is valid too.

Theorem 2 (Converse of Point on angle bisector theorem)

If a point is equidistant from the sides of an angle, then it is located on an angle bisector.

<TABLE>
 <TR>
 <TD>
Proof

For your convenience the Figure 1 of the previous solution is shown one more&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
time below as the Figure 2. This time we are given the angle BAC and 
the point P inside it, which is equidistant from the angle sides. This 
means that the perpendiculars PD and PE drawn from this point to the 
angle sides are of equal length. We need to prove that the ray AP bisects 
the angle BAC in the two congruent angles CAP and PAB.

Draw the straight segment DE connecting the points D and E (Figure 3).
Since the segments PD and PE are of equal length, the triangle DPE 
is isosceles. Therefore, its angles PDE and PED are congruent as the angles
at the base of the isosceles triangle (see the lesson <A HREF=http://www.algebra.com/algebra/homework/Triangles/Isosceles-triangles.lesson>Isosceles triangles</A> 
under the current topic Triangles in the section Geometry in this site).
This implies that the angles ADE and AED are congruent as they 
complement the angles PDE and PED to the right angles PDA and PEA. 
 </TD>
 <TD>
{{{drawing( 200, 200, 0, 4, 0, 4, 
 line( 0.3, 0.5, 3.7, 0.5), 
 line( 0.3, 0.5, 2.0, 3.5),
 red(line( 0.3, 0.5, 3.7, 2.5)),

circle(2.5, 1.8, 0.05, 0.05),

locate(0.3, 0.5, A),
            locate(3.7, 0.5, B),
            locate(2.0, 3.9, C),
            locate(2.5, 2.2, P),

green(line (2.5, 1.8, 2.5, 0.5)),
            locate(2.5, 0.5, D),

green(line (2.5, 1.8, 1.4, 2.45)),
 locate(1.2, 2.8, E)
)}}}
Figure 2. To the Theorem 2&nbsp;&nbsp;&nbsp;&nbsp;
 </TD>
 <TD>
{{{drawing( 200, 200, 0, 4, 0, 4, 
 line( 0.3, 0.5, 3.7, 0.5), 
 line( 0.3, 0.5, 2.0, 3.5),
 red(line( 0.3, 0.5, 3.7, 2.5)),

circle(2.5, 1.8, 0.05, 0.05),

locate(0.3, 0.5, A),
            locate(3.7, 0.5, B),
            locate(2.0, 3.9, C),
            locate(2.5, 2.2, P),

green(line (2.5, 1.8, 2.5, 0.5)),
            locate(2.5, 0.5, D),

green(line (2.5, 1.8, 1.4, 2.45)),
            locate(1.2, 2.8, E),

green(line (1.4, 2.45, 2.5, 0.5)),

arc(1.4, 2.45, 0.8, 0.8,  30,  63),

arc(2.5, 0.5,  0.8, 0.8, 237, 270),

arc(1.4, 2.45, 0.9, 0.9,  63, 120),
            arc(1.4, 2.45, 1.0, 1.0,  63, 120),

arc(2.5, 0.5,  0.9, 0.9, 180, 237),
            arc(2.5, 0.5,  1.0, 1.0, 180, 237),

arc(0.3, 0.5, 0.8, 0.8, 300, 330),
            arc(0.3, 0.5, 0.9, 0.9, 300, 330),
            arc(0.3, 0.5, 1.0, 1.0, 300, 330),

arc(0.3, 0.5, 1.0, 1.0, 330, 360),
 arc(0.3, 0.5, 1.1, 1.1, 330, 360),
 arc(0.3, 0.5, 1.2, 1.2, 330, 360)
)}}}
Figure 3. To the proof of the Theorem 2
 </TD>
 </TR>
</TABLE>

The fact that the angles ADE and AED are congruent means that the triangle ADE is isosceles: AD = AE (see the same lesson <A HREF=http://www.algebra.com/algebra/homework/Triangles/Isosceles-triangles.lesson>Isosceles triangles</A> again).
Thus, the triangles ADP and AEP have the congruent sides AD = AE, PD = PE and the common side AP. 
Hence, these triangles are congruent in accordance with the postulate 3 (SSS) of the lesson <A HREF=http://www.algebra.com/algebra/homework/Triangles/Congruence-tests-for-triangles.lesson>Congruence_tests_for_triangles</A> of the current topic Triangles in 
the section Geometry in this site.
Therefore, the angles BAP and PAC are congruent as the corresponding angles of the congruent triangles ADP and AEP. 
This is what to be demonstrated. The proof is completed.

Summary

A point is equidistant from the sides of an angle if and only if the point lies on the angle bisector.

We can present this result in the following terms:

&nbsp;&nbsp;&nbsp;&nbsp;- the set of points equidistant from the sides of an angle is the angle bisector, or
 &nbsp;&nbsp;&nbsp;&nbsp;- an angle bisector is the set of points equidistant from the sides of an angle.

<BLOCKQUOTE>Reminder

In the lesson <A HREF=http://www.algebra.com/algebra/homework/Triangles/A-perpendicular-bisector-of-a-segment.lesson>A perpendicular bisector of a segment</A> under the current topic Triangles in the section Geometry in this site we defined the Geometry term locus as the set of points satisfying some certain condition. It was shown in that lesson that the perpendicular drawn through the midpoint of a line segment is the locus of points equidistant from the endpoints of the segment</BLOCKQUOTE>

Based on the results obtained in this lesson, one can say that 
 &nbsp;&nbsp;&nbsp;&nbsp;- an angle bisector is the locus of points equidistant from the sides of the angle, &nbsp;or 
 &nbsp;&nbsp;&nbsp;&nbsp;- the locus of points equidistant from the sides of an angle is the angle bisector.

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>.