Is the angle bisector and median of an isosceles triangle same?

Newland Z.

asked • 11/04/14

If a median of a triangle is also an angle bisector, is it also an altitude? I need help with this

1 Expert Answer

If a median is also an angle bisector, then  the triangle is an isosceles triangle. 

The proof is this statement is a bit tricky.   Call the base angles   Θ1  and Θ2 .     The law of sines can be used to show that   sin(Θ1) =  sin(Θ2).     There are two possibilities:    1)   Θ1 =  Θ2  or  2)  Θ1 =  180 - Θ2 .   However, 2) can be ruled out because it implies that Θ1 +  Θ2  =  180 which forces the vertex angle to be zero measure.   This leaves Θ1 = Θ2 , which means that the triangle is isosceles. 

The median divides an isosceles triangle into two congruent triangles by sss.   The two angles made by the intersection of the median and the base are equal measure by corresponding parts.  However their sum is 180 degrees.  Thus both of these angles must be right angles.  

Since the median meets the base at right angles, it is an altitude.

Lets start by defining them:  median, angle bisector and altitude.
Median – A line segment joining a vertex of a triangle with the mid-point of the opposite side.

Angle Bisector – A line segment joining a vertex of a triangle with the opposite side such that the angle at the vertex is split into two equal parts.

Altitude – A line segment joining a vertex of a triangle with the opposite side such that the segment is perpendicular to the opposite side.

Usually, medians, angle bisectors and altitudes drawn from the same vertex of a triangle are different line segments. But in special triangles such as isosceles and equilateral, they can overlap. We will now give you some properties which can be very useful.

I.

In an isosceles triangle (where base is the side which is not equal to any other side):

– the altitude drawn to the base is the median and the angle bisector;

– the median drawn to the base is the altitude and the angle bisector;

– the bisector of the angle opposite to the base is the altitude and the median.

II.

The reverse is also true. Consider a triangle ABC:

– If angle bisector of vertex A is also the median, the triangle is isosceles such that AB = AC and BC is the base. Hence this angle bisector is also the altitude.

– If altitude drawn from vertex A is also the median, the triangle is isosceles such that AB = AC and BC is the base. Hence this altitude is also the angle bisector.

– If median drawn from vertex A is also the angle bisector, the triangle is isosceles such that AB = AC and BC is the base. Hence this median is also the altitude.

and so on…

III.

In an equilateral triangle, each altitude, median and angle bisector drawn from the same vertex, overlap.

Try to prove all these properties on your own. That way, you will not forget them.

A few things this implies:

–          Should an angle bisector in a triangle which is also a median be perpendicular to the opposite side? Yes.

–          Can we have an angle bisector which is also a median which is not perpendicular? No. Angle bisector which is also a median implies isosceles triangle which implies it is also the altitude.

–          Can we have a median from vertex A which is perpendicular to BC but does not bisect the angle A? No. A median which is an altitude implies the triangle is isosceles which implies it is also the angle bisector.

and so on…

Let’s take a quick question on these concepts:

Question: What is ?A in triangle ABC?

Statement 1: The bisector of ?A is a median in triangle ABC.
Statement 2: The altitude of B to AC is a median in triangle ABC.

Solution: We are given a triangle ABC but we don’t know what kind of a triangle it is.

Jump on to the statements directly.

Statement 1: The bisector of ?A is a median in triangle ABC.

The angle bisector is also a median. This means triangle ABC must be an isosceles triangle such that AB = AC. But we have no idea about the measure of angle A. This statement alone is not sufficient.

Statement 2: The altitude of ?B to AC is a median in triangle ABC.

The altitude is also a median. This means triangle ABC must be an isosceles triangle such that AB = BC (Note that the altitude is drawn from vertex B here). But we have no idea about the measure of angle A. This statement alone is not sufficient.

Using both statements together, we see that AB = AC = BC. So the triangle is equilateral! So angle A must be 60 degrees. Sufficient!

Answer (C)

This Lesson (An altitude a median and an angle bisector in the isosceles triangle) was created by by ikleyn(46438)

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Theorem 1
In an isosceles triangle the altitude drawn to the base is the median and the angle bisector.

Proof Let ABC be an isosceles triangle with sides AC and BC of equal length (Figure 1). The segment CD is an altitude drawn to the base AB of the triangle. We need to prove that CD is the median of the triangle ABC and the angle bisector of the angle ACB opposite to the base. Since CD is an altitude, the triangles ADC and BDC are right triangles. The angles BAC and ABC are congruent as the angles at the base of the isosceles triangle ABC               (this was proved in the lesson Isosceles triangles under the current topic in this site). Therefore, the triangles ADC and BDC are right triangles with congruent angles DAC and DBC. This implies that the angles ACD and BCD are congruent as complementary angles to the angles DAC and DBC. It means that the altitude CD is the bisector of the angle ACB. Thus, this part is proved. This also means the triangles ADC and BDC have two pairs of congruent angles ACD = BCD and DAC = DBC that include congruent sides AC = BC. Hence, these triangles are congruent, in accordance to the postulate P2 (ASA) of the triangle congruency (see the lesson Congruence tests for triangles under the current topic in this site).

Figure 1. To the Theorem 1

Therefore, the corresponding sides are congruent: AD = BD, and CD is the median of the triangle ABC. The proof is completed. It is not difficult to prove, in the same manner, two symmetrical statements:     - In an isosceles triangle the median drawn to the base is the altitude and the angle bisector.     - In an isosceles triangle the bisector of the angle opposite to the base is the altitude and the median. Below are two independent proofs.

Theorem 2
In an isosceles triangle the median drawn to the base is the altitude and the angle bisector.

Proof Figure 2 shows an isosceles triangle ABC with sides AC and BC of equal length. The segment CD is the median drawn to the base AB of the triangle. We need to prove that CD is the altitude of the triangle ABC and the angle bisector of the angle ACB opposite to the base. Since CD is a median, the segments AD and BD are congruent. Thus, the triangles ADC and BDC have two pairs of congruent sides AD = BD, AC = BC and                    the common side CD. Hence, these triangles are congruent, in accordance to the postulate P3 (SSS) of the triangle congruency (see the lesson Congruence tests for triangles under the current topic in this site). It implies that the angles ACD and BCD are congruent. This means that the median CD is the bisector of the angle ACB. Thus, this part is proved. This also means that the corresponding angles ADC and BDC are congruent. Since these two angles give in sum the straight angle of 180�, each of them is 90�, that is the right angle. Thus, the median CD is the altitude of the triangle ABC.

Figure 2. To the Theorem 2

The proof is completed.

Theorem 3
In an isosceles triangle the bisector of the angle opposite to the base is the altitude and the median.

Proof Figure 3 shows an isosceles triangle ABC with sides AC and BC of equal length. The segment CD is the bisector of the angle ACB opposite to the base AC. We need to prove that CD is the altitude and the median of the triangle ABC. Since CD is an angle ACB bisector, the angles ACD and BCD are congruent. Thus, the triangles ADC and BDC have the pair of congruent sides AD = BD, the common                       side CD and the congruent included angles ACD and BCD. Hence, these triangles are congruent, in accordance to the postulate P1 (SAS) of the triangle congruency (see the lesson Congruence tests for triangles under the current topic in this site). It implies that the segments AD and BD are congruent. This means that the bisector of the angle ACB is the median. Thus, this part is proved. This also means that the corresponding angles ADC and BDC are congruent. Since these two angles give in sum the straight angle of 180�, each of them is 90�, that is the right angle. Thus, the bisector of the angle ACB is the altitude of the triangle ABC.

Figure 3. To the Theorem 3

The proof is completed.

Summary

In an isosceles triangle
  - the altitude drawn to the base is the median and the angle bisector;
  - the median drawn to the base is the altitude and the angle bisector;
  - the bisector of the angle opposite to the base is the altitude and the median. In an isosceles triangle the altitudes,  the medians and the angle bisectors drawn to the lateral sides from the angles at the base also have remarkable properties. They are considered in the lessons

Altitudes in an isosceles triangle,

Medians in an isosceles triangle,  and
Angle bisectors in an isosceles triangle  under the topic  Geometry  of the section  Word problems  in this site. For your convenience,  below is the list of my relevant lessons in this site in the logical order.

Congruence tests for triangles                                                       under the topic Triangles in the section Geometry;

Isosceles triangles                                                                        under the topic Triangles in the section Geometry.

For navigation over the lessons on Properties of Triangles use this file/link  Properties of Trianles.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link  GEOMETRY - YOUR ONLINE TEXTBOOK.

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