Lesson Explainer: Corollaries of Isosceles Triangle Theorems | Nagwa Lesson Explainer: Corollaries of Isosceles Triangle Theorems | Nagwa

Lesson Explainer: Corollaries of Isosceles Triangle Theorems Mathematics

In this explainer, we will learn how to use the corollaries of the isosceles triangle theorems to find missing lengths and angles in isosceles triangles.

Let’s begin by recapping the exact definition of an isosceles triangle. Recall that congruent means having the same measure; for example, congruent sides have the same length and congruent angles have the same measure.

Definition: Isosceles Triangle

An isosceles triangle is a triangle that has two congruent sides.

The congruent sides are called the legs of the triangle, and the third side is called the base.

Because isosceles triangles have two congruent sides, this leads us to an important angle property of isosceles triangles.

Theorem: Isosceles Triangle Theorem

If two sides of a triangle are congruent, then the angles opposite those sides are congruent.

The congruent angles are called the base angles. The third angle is called the vertex angle.

In 𝐴𝐵𝐶, 𝐵𝐴=𝐵𝐶, so 𝑚𝐴=𝑚𝐶.

We know that isosceles triangles, by definition, have two congruent sides, and by the previous theorem, they have two congruent angles. The converse of this is true; that is, if a triangle has two angles that are congruent, then it is an isosceles triangle. This is defined in the theorem below.

Theorem: Converse of the Isosceles Triangle Theorem

If two angles of a triangle are congruent, the sides opposite those angles are also congruent.

In this explainer, we will consider a number of corollaries to these theorems. These corollaries allow us to identify additional geometric properties about isosceles triangles. Let’s see the first of these corollaries.

Corollary of the Isosceles Triangle Theorems: The Median of an Isosceles Triangle

The median of an isosceles triangle from the vertex angle bisects it and is perpendicular to the base.

We can prove this corollary as follows.

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side and thus bisecting that side. In isosceles triangle 𝐴𝐵𝐶 below, the median from the vertex angle, 𝐴, is a line segment joining 𝐴 to the midpoint (𝐷) of the base, 𝐵𝐶.

Since 𝐷 is the midpoint of 𝐵𝐶, we know that 𝐶𝐷=𝐵𝐷.

We can also write that 𝐴𝐶=𝐴𝐵,(sincethetriangleisisosceles) and 𝐴𝐷=𝐴𝐷.(acommonside)

As there are three congruent sides, then 𝐴𝐷𝐶 is congruent to 𝐴𝐷𝐵 by the SSS congruency criterion.

Hence, we can write that 𝑚𝐴𝐷𝐶=𝑚𝐴𝐷𝐵, and since 𝐵𝐶 is a straight line, these angles must both be 90. We can also note that 𝑚𝐶𝐴𝐷=𝑚𝐵𝐴𝐷, so the median from the vertex of an isosceles triangle also bisects the vertex.

Given that 𝐴𝐷 is a median, we can say that the median from the vertex angle bisects the base, but here we have also demonstrated that it does this at an angle of 90.

Therefore, the median of an isosceles triangle from the vertex angle is a perpendicular bisector of the triangle.

Because of this corollary, we can observe that a useful property of the median of an isosceles triangle is that it also forms the axis of symmetry of an isosceles triangle and splits the isosceles triangle into two congruent right triangles.

We will now see how we can apply this corollary in the following example.

Example 1: Finding Values in a Triangle given a Perpendicular Bisector

For which values of 𝑥 and 𝑦 is 𝐴𝐷 a perpendicular bisector of 𝐵𝐶?

Answer

In the figure, we can observe that 𝐴𝐵𝐶 appears to be an isosceles triangle. Although we cannot prove this, we can use some of the properties of isosceles triangles to help. An isosceles triangle is a triangle that has two congruent sides. We recall that the median of an isosceles triangle from the vertex angle is a perpendicular bisector of the base.

Hence, 𝐴𝐷 is only a perpendicular bisector of 𝐵𝐶 in the case of an isosceles triangle. Therefore, we need to determine the values of 𝑥 and 𝑦 such that 𝐴𝐵=𝐴𝐶 (two congruent legs) and 𝐵𝐷=𝐶𝐷 (the base is bisected).

We can begin by substituting in the given expressions for 𝐴𝐵 and 𝐴𝐶. This gives 𝐴𝐵=𝐴𝐶3𝑥+2=5𝑦+33𝑥1=5𝑦.

In the same way, for 𝐵𝐷 and 𝐶𝐷, we can form a second equation as follows: 𝐵𝐷=𝐶𝐷5𝑦1=103𝑥5𝑦=113𝑥.

To find the values of 𝑥 and 𝑦, we solve these two equations simultaneously, either by substitution or elimination. They are:

3𝑥1=5𝑦,113𝑥=5𝑦.(1)(2)

As both equations have a term of 5𝑦 on the right-hand side of each equation, we can set the left-hand side of each equation equal to one another and solve for 𝑥. This gives us 3𝑥1=113𝑥6𝑥1=116𝑥=12𝑥=2.

Next, we substitute 𝑥=2 into either equation (1) or (2). Substituting into equation (1) gives 3(2)1=5𝑦61=5𝑦5=5𝑦𝑦=1.

Thus, we can answer that 𝐴𝐷 is a perpendicular bisector of 𝐵𝐶 when 𝑥=2 and 𝑦=1.

Let’s consider another corollary.

Corollary of the Isosceles Triangle Theorems: Bisector of the Vertex Angle

The bisector of the vertex angle of an isosceles triangle is a perpendicular bisector of the base.

We can prove this in the following way. Consider the isosceles triangle 𝐸𝐹𝐺. The bisector of the vertex angle, 𝐸, has been drawn.

We can label the point where the bisector crosses 𝐺𝐹 as 𝐻. Since 𝐸𝐻 is an angle bisector, we know that 𝑚𝐺𝐸𝐻=𝑚𝐹𝐸𝐻.

There are now 2 pairs of congruent sides: 𝐸𝐺=𝐸𝐹𝐸𝐹𝐺𝐸𝐻.(sinceisisosceles)andisacommonside

With the congruent pair of included angles, we can say that 𝐸𝐺𝐻 is congruent to 𝐸𝐹𝐻 by the SAS condition (two equal sides and the included angle). You may be familiar with this criterion although we will not directly prove it in this explainer.

This means that corresponding angles are congruent and 𝑚𝐸𝐻𝐺=𝑚𝐸𝐻𝐹. Since these two angles lie on the straight line 𝐺𝐹, then they must both equal 90. Furthermore, the sides 𝐺𝐻 and 𝐹𝐻 are corresponding, and so they are also congruent.

Thus, we have proved the corollary that the bisector of the vertex angle 𝐸 is also the perpendicular bisector of the isosceles triangle 𝐸𝐹𝐺.

Let’s see how we can apply this in the following example.

Example 2: Finding a Missing Length in an Isosceles Triangle Using the Bisector of the Vertex Angle

  1. Fill in the blank: In this figure, if 𝐴𝐵=𝐴𝐶, 𝐴𝐷𝐵𝐶={𝐷}, where 𝐷𝐶=8cm, and 𝑚𝐶𝐴𝐷=𝑚𝐵𝐴𝐷=35, the length of 𝐵𝐷 is cm.
  2. Find 𝑚𝐴𝐶𝐵.

Answer

For the figure, we are given the information that 𝐴𝐵𝐶 has two congruent sides (𝐴𝐵=𝐴𝐶). This means that 𝐴𝐵𝐶 is an isosceles triangle. We are also given that the intersection point of 𝐴𝐷 and 𝐵𝐶 is the point 𝐷.

Part 1

Let’s begin by adding the angle measurements, 𝑚𝐶𝐴𝐷=𝑚𝐵𝐴𝐷=35, to the figure along with marking the congruent sides.

We observe that, since these two angles are equal in measure, this means that the vertex angle, 𝐴, of the isosceles triangle has been bisected. We recall that one of the corollaries of the isosceles triangle theorems tells us that the bisector of the vertex angle of an isosceles triangle is a perpendicular bisector of the base. This allows us to calculate the length of 𝐵𝐷. Note that, if we did not have this information about the angles, we would not have been able to prove that 𝐴𝐷 is a perpendicular bisector.

Since 𝐵𝐶 has been bisected, then we know that 𝐵𝐷=𝐷𝐶.

From the diagram, we have that 𝐷𝐶=8cm. Hence, 𝐵𝐷=8.cm

Therefore, we can fill in the blank: the length of 𝐵𝐷 is 8 cm.

Part 2

To find 𝑚𝐴𝐶𝐵, we can use the fact that 𝐴𝐷 is a perpendicular bisector of 𝐴𝐵𝐶. This means that 𝑚𝐴𝐷𝐶=90.

Using 𝐴𝐶𝐷 and the fact that the interior angle measures in a triangle sum to 180, we have 𝑚𝐴𝐶𝐷+𝑚𝐶𝐴𝐷+𝑚𝐴𝐷𝐶=180𝑚𝐴𝐶𝐷+35+90=180𝑚𝐴𝐶𝐷+125=180𝑚𝐴𝐶𝐷=180125=55.

Since 𝐴𝐶𝐷 is the same angle as 𝐴𝐶𝐵, we can give the answer that 𝑚𝐴𝐶𝐵 is 55.

We will now see another corollary of the isosceles triangle theorem.

Corollary of the Isosceles Triangle Theorems: Perpendicular to the Base

The straight line that passes through the vertex angle of an isosceles triangle and is perpendicular to the base bisects the base and the vertex angle.

We can prove this by considering an example of a straight line passing through the vertex angle of an isosceles triangle, 𝑃𝑄𝑅 below, at 90 to the base. We can define the point at which the line intersects 𝑅𝑄 as 𝑇.

At this point, we need to establish how this line affects the vertex angle 𝑃 and the base 𝑅𝑄.

Given that the line is perpendicular to the base, we recognize that 𝑚𝑃𝑇𝑅=𝑚𝑃𝑇𝑄=90.

As the triangle is isosceles, we know that 𝑅𝑃=𝑄𝑃.

We can observe that 𝑅𝑃 and 𝑄𝑃 are the hypotenuses of their respective right triangles, 𝑃𝑅𝑇 and 𝑃𝑄𝑇.

Furthermore, since 𝑃𝑇 is a common side to both 𝑃𝑅𝑇 and 𝑃𝑄𝑇, we know that this a congruent side in these triangles. Therefore, by applying the RHS (right angle-hypotenuse-side) congruence criterion, we can prove that 𝑃𝑅𝑇𝑃𝑄𝑇.

The corollary statement can then be verified: the base 𝑅𝑄 is bisected, and since 𝑅𝑇 and 𝑄𝑇 are the corresponding sides of two congruent triangles, the sides themselves are also congruent. This means that the base 𝑅𝑄 is bisected. The vertex angle 𝑃 has also been bisected, as we know that its two component angles, 𝑅𝑃𝑇 and 𝑄𝑃𝑇, are the corresponding angles of two congruent triangles and, hence, have equal measures.

We will now see how we can apply this corollary in the following example.

Example 3: Finding the Measure of an Angle in an Isosceles Triangle Using Its Properties

Find 𝑚𝐷𝐴𝐵.

Answer

In the figure, we observe that the largest triangle, 𝐴𝐵𝐶, has two congruent sides marked to show us that 𝐴𝐶=𝐴𝐵. Therefore, 𝐴𝐵𝐶 must be an isosceles triangle.

We are given that 𝑚𝐴𝐷𝐵=90. This means that the line segment, 𝐴𝐷, from the vertex angle in the isosceles triangle is perpendicular to the base. We recall that one of the corollaries to the isosceles triangle theorems states that the straight line that passes through the vertex angle of an isosceles triangle and is perpendicular to the base bisects the base and the vertex angle. This is useful in allowing us to determine 𝑚𝐷𝐴𝐵.

Since 𝐶𝐴𝐵 has been bisected, then 𝑚𝐷𝐴𝐵=𝑚𝐶𝐴𝐷.

From the diagram, we have 𝑚𝐶𝐴𝐷=25; hence, we can determine that the answer is 𝑚𝐷𝐴𝐵=25.

We will now see two examples where we need to apply several of these corollaries to the isosceles triangle theorems to find the measures of unknown angles and sides.

Example 4: Proving a Geometric Statement Using the Corollaries of the Isosceles Triangle Theorems

In the following figure, if 𝐿𝑋=𝑌𝑋, 𝐿𝑍=𝑌𝑍, 𝑚𝐿𝑌𝑋=30, 𝑚𝑍𝐿𝑌=55, and 𝑀 is the midpoint of 𝐿𝑌, find 𝑚𝑀𝑍𝐿 and 𝑚𝐿𝑀𝑋.

Answer

We are given the information that 𝐿𝑋=𝑌𝑋 and 𝐿𝑍=𝑌𝑍. This means that, in each of 𝐿𝑍𝑌 and 𝐿𝑋𝑌, there is a pair of congruent sides. A triangle with two congruent sides is, by definition, an isosceles triangle. Therefore, 𝐿𝑍𝑌 and 𝐿𝑋𝑌 are both isosceles. Knowing the properties of isosceles triangles given in the corollaries of the isosceles triangle theorems will be useful in calculating the required angles.

Let’s begin by constructing the segment 𝑍𝑀, which creates the first required angle, 𝑀𝑍𝐿.

As 𝑀 is the midpoint of 𝐿𝑌, we know that 𝑍𝑀 must be a median of the isosceles triangle 𝐿𝑍𝑌. We recall that the median of an isosceles triangle from the vertex angle is a perpendicular bisector of the base. As 𝑍𝑀 is a perpendicular bisector to 𝐿𝑌, then 𝑚𝐿𝑀𝑍=90.

Within 𝐿𝑀𝑍, we can use the property that the internal angle measures in a triangle sum to 180 to help us find 𝑚𝑀𝑍𝐿. Given that 𝑚𝑍𝐿𝑀=55 and 𝑚𝐿𝑀𝑍=90, we have 𝑚𝑍𝐿𝑀+𝑚𝐿𝑀𝑍+𝑚𝑀𝑍𝐿=18055+90+𝑚𝑀𝑍𝐿=180145+𝑚𝑀𝑍𝐿=180𝑚𝑀𝑍𝐿=180145=35.

Next, we must determine 𝑚𝐿𝑀𝑋. We construct the segment 𝑀𝑋 to create this angle.

In the same process as before, we know that 𝐿𝑋𝑌 is isosceles. Since 𝑀 is a midpoint of 𝐿𝑌, then the median from the vertex angle 𝑋 is a perpendicular bisector of the base. This means that 𝑚𝐿𝑀𝑋=90.

We can give the answers for 𝑚𝑀𝑍𝐿 and 𝑚𝐿𝑀𝑋 as 3590.and

We will now see one final example.

Example 5: Proving a Geometric Statement Using the Corollaries of the Isosceles Triangle Theorems

In the figure below, 𝐴𝐶𝐵𝐷=𝑀, 𝑀𝐷=𝑀𝐶, 𝐴𝐵𝐶𝐷, and 𝑚𝑀𝐶𝐷=40. Draw 𝑀𝐸𝐶𝐷 and cut 𝐶𝐷 at 𝐸, then draw 𝑀𝐹𝐴𝐵 and cut 𝐴𝐵 at 𝐹. Find 𝑚𝐶𝑀𝐸 and 𝑚𝐵𝑀𝐹.

Answer

We can begin by constructing the two segments 𝑀𝐸 and 𝑀𝐹, which are perpendicular to 𝐶𝐷 and 𝐴𝐵 respectively. We can also fill in the given angle information, which is 𝑚𝑀𝐶𝐷=40, and mark on the diagram that 𝑀𝐷=𝑀𝐶.

In the lower triangle, 𝐶𝑀𝐷, we have two congruent line segments. This means that 𝐶𝑀𝐷 must be an isosceles triangle. There are a number of ways in which we can find 𝑚𝐶𝑀𝐸. One method is to use 𝐶𝑀𝐸 and recall that the internal angle measures in a triangle sum to 180. Given that 𝑚𝑀𝐶𝐸=40 and 𝑚𝐶𝐸𝑀=90, we have 𝑚𝑀𝐶𝐸+𝑚𝐶𝐸𝑀+𝑚𝐶𝑀𝐸=18040+90+𝑚𝐶𝑀𝐸=180130+𝑚𝐶𝑀𝐸=180𝑚𝐶𝑀𝐸=180130=50.

Next, we need to calculate the required angle, 𝑚𝐵𝑀𝐹. Given that 𝐶𝑀𝐷 is an isosceles triangle, we can use the fact that the straight line that passes through the vertex angle of an isosceles triangle and is perpendicular to the base bisects the base and the vertex angle. This means that 𝑚𝐷𝑀𝐸=𝑚𝐶𝑀𝐸=50.

Given that 𝐴𝐵𝐶𝐷, 𝑀𝐸𝐶𝐷, and 𝑀𝐹𝐴𝐵 with 𝑀𝐸 and 𝑀𝐹 meeting at point 𝑀, then 𝐸𝐹 is a straight line. We can then observe that 𝐷𝑀𝐸 and 𝐵𝑀𝐹 are vertically opposite and, hence, are congruent.

Therefore, we can give the answer that the unknown angle measures are 𝑚𝐶𝑀𝐸=50𝑚𝐵𝑀𝐹=50.and

We now summarize the key points.

Key Points

  • An isosceles triangle is a triangle that has two congruent sides.
  • If two sides of a triangle are congruent, then the angles opposite those sides are congruent.
  • If two angles of a triangle are congruent, then the sides opposite those angles are also congruent.
  • The median of an isosceles triangle from the vertex angle is a perpendicular bisector of the base and bisects the vertex angle.
  • The bisector of the vertex angle of an isosceles triangle is a perpendicular bisector of the base.
  • The straight line that passes through the vertex angle of an isosceles triangle and is perpendicular to the base bisects the base and the vertex angle.
  • The axis of symmetry of an isosceles triangle is the median that bisects the base.

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