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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 △𝐴𝐡𝐢, 𝐡𝐴=𝐡𝐢, so π‘šβˆ π΄=π‘šβˆ πΆ.

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 𝐴𝐡𝐢 below, the median from the vertex angle, 𝐴, is a line segment joining 𝐴 to the midpoint (𝐷) of the base, 𝐡𝐢.

Since 𝐷 is the midpoint of 𝐡𝐢, we know that 𝐢𝐷=𝐡𝐷.

We can also write that 𝐴𝐢=𝐴𝐡,(sincethetriangleisisosceles) and 𝐴𝐷=𝐴𝐷.(acommonside)

As there are three congruent sides, then △𝐴𝐷𝐢 is congruent to △𝐴𝐷𝐡 by the SSS congruency criterion.

Hence, we can write that π‘šβˆ π΄π·πΆ=π‘šβˆ π΄π·π΅, and since 𝐡𝐢 is a straight line, these angles must both be 90∘. We can also note that π‘šβˆ πΆπ΄π·=π‘šβˆ π΅π΄π·, 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 π‘₯ and 𝑦 is 𝐴𝐷 a perpendicular bisector of 𝐡𝐢?

Answer

In the figure, we can observe that △𝐴𝐡𝐢 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 𝐡𝐢 in the case of an isosceles triangle. Therefore, we need to determine the values of π‘₯ and 𝑦 such that 𝐴𝐡=𝐴𝐢 (two congruent legs) and 𝐡𝐷=𝐢𝐷 (the base is bisected).

We can begin by substituting in the given expressions for 𝐴𝐡 and 𝐴𝐢. This gives 𝐴𝐡=𝐴𝐢3π‘₯+2=5𝑦+33π‘₯βˆ’1=5𝑦.

In the same way, for 𝐡𝐷 and 𝐢𝐷, we can form a second equation as follows: 𝐡𝐷=𝐢𝐷5π‘¦βˆ’1=10βˆ’3π‘₯5𝑦=11βˆ’3π‘₯.

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

3π‘₯βˆ’1=5𝑦,11βˆ’3π‘₯=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 π‘₯. This gives us 3π‘₯βˆ’1=11βˆ’3π‘₯6π‘₯βˆ’1=116π‘₯=12π‘₯=2.

Next, we substitute π‘₯=2 into either equation (1) or (2). Substituting into equation (1) gives 3(2)βˆ’1=5𝑦6βˆ’1=5𝑦5=5𝑦𝑦=1.

Thus, we can answer that 𝐴𝐷 is a perpendicular bisector of 𝐡𝐢 when π‘₯=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 π‘šβˆ πΊπΈπ»=π‘šβˆ πΉπΈπ».

There are now 2 pairs of congruent sides: 𝐸𝐺=𝐸𝐹△𝐸𝐹𝐺𝐸𝐻.(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 π‘šβˆ πΈπ»πΊ=π‘šβˆ πΈπ»πΉ. 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 𝐴𝐡=𝐴𝐢, 𝐴𝐷∩𝐡𝐢={𝐷}, where 𝐷𝐢=8cm, and π‘šβˆ πΆπ΄π·=π‘šβˆ π΅π΄π·=35∘, the length of 𝐡𝐷 is cm.
  2. Find π‘šβˆ π΄πΆπ΅.

Answer

For the figure, we are given the information that △𝐴𝐡𝐢 has two congruent sides (𝐴𝐡=𝐴𝐢). This means that △𝐴𝐡𝐢 is an isosceles triangle. We are also given that the intersection point of 𝐴𝐷 and 𝐡𝐢 is the point 𝐷.

Part 1

Let’s begin by adding the angle measurements, π‘šβˆ πΆπ΄π·=π‘šβˆ π΅π΄π·=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 𝐡𝐷. 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 𝐡𝐢 has been bisected, then we know that 𝐡𝐷=𝐷𝐢.

From the diagram, we have that 𝐷𝐢=8cm. Hence, 𝐡𝐷=8.cm

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

Part 2

To find π‘šβˆ π΄πΆπ΅, we can use the fact that 𝐴𝐷 is a perpendicular bisector of △𝐴𝐡𝐢. This means that π‘šβˆ π΄π·πΆ=90∘.

Using △𝐴𝐢𝐷 and the fact that the interior angle measures in a triangle sum to 180∘, we have π‘šβˆ π΄πΆπ·+π‘šβˆ πΆπ΄π·+π‘šβˆ π΄π·πΆ=180π‘šβˆ π΄πΆπ·+35+90=180π‘šβˆ π΄πΆπ·+125=180π‘šβˆ π΄πΆπ·=180βˆ’125=55.∘∘∘∘∘∘∘∘∘

Since ∠𝐴𝐢𝐷 is the same angle as ∠𝐴𝐢𝐡, we can give the answer that π‘šβˆ π΄πΆπ΅ 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 π‘šβˆ π‘ƒπ‘‡π‘…=π‘šβˆ π‘ƒπ‘‡π‘„=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 △𝑃𝑅𝑇≅△𝑃𝑄𝑇.

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, βˆ π‘…π‘ƒπ‘‡ and βˆ π‘„π‘ƒπ‘‡, 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 π‘šβˆ π·π΄π΅.

Answer

In the figure, we observe that the largest triangle, △𝐴𝐡𝐢, has two congruent sides marked to show us that 𝐴𝐢=𝐴𝐡. Therefore, △𝐴𝐡𝐢 must be an isosceles triangle.

We are given that π‘šβˆ π΄π·π΅=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 π‘šβˆ π·π΄π΅.

Since ∠𝐢𝐴𝐡 has been bisected, then π‘šβˆ π·π΄π΅=π‘šβˆ πΆπ΄π·.

From the diagram, we have π‘šβˆ πΆπ΄π·=25∘; hence, we can determine that the answer is π‘šβˆ π·π΄π΅=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=π‘Œπ‘‹, 𝐿𝑍=π‘Œπ‘, π‘šβˆ πΏπ‘Œπ‘‹=30∘, π‘šβˆ π‘πΏπ‘Œ=55∘, and 𝑀 is the midpoint of πΏπ‘Œ, find π‘šβˆ π‘€π‘πΏ and π‘šβˆ πΏπ‘€π‘‹.

Answer

We are given the information that 𝐿𝑋=π‘Œπ‘‹ and 𝐿𝑍=π‘Œπ‘. This means that, in each of β–³πΏπ‘π‘Œ and β–³πΏπ‘‹π‘Œ, there is a pair of congruent sides. A triangle with two congruent sides is, by definition, an isosceles triangle. Therefore, β–³πΏπ‘π‘Œ and β–³πΏπ‘‹π‘Œ 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, βˆ π‘€π‘πΏ.

As 𝑀 is the midpoint of πΏπ‘Œ, we know that 𝑍𝑀 must be a median of the isosceles triangle πΏπ‘π‘Œ. 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 πΏπ‘Œ, then π‘šβˆ πΏπ‘€π‘=90.∘

Within △𝐿𝑀𝑍, we can use the property that the internal angle measures in a triangle sum to 180∘ to help us find π‘šβˆ π‘€π‘πΏ. Given that π‘šβˆ π‘πΏπ‘€=55∘ and π‘šβˆ πΏπ‘€π‘=90∘, we have π‘šβˆ π‘πΏπ‘€+π‘šβˆ πΏπ‘€π‘+π‘šβˆ π‘€π‘πΏ=18055+90+π‘šβˆ π‘€π‘πΏ=180145+π‘šβˆ π‘€π‘πΏ=180π‘šβˆ π‘€π‘πΏ=180βˆ’145=35.∘∘∘∘∘∘∘∘∘

Next, we must determine π‘šβˆ πΏπ‘€π‘‹. We construct the segment 𝑀𝑋 to create this angle.

In the same process as before, we know that β–³πΏπ‘‹π‘Œ is isosceles. Since 𝑀 is a midpoint of πΏπ‘Œ, then the median from the vertex angle 𝑋 is a perpendicular bisector of the base. This means that π‘šβˆ πΏπ‘€π‘‹=90.∘

We can give the answers for π‘šβˆ π‘€π‘πΏ and π‘šβˆ πΏπ‘€π‘‹ 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, 𝐴𝐢∩𝐡𝐷=𝑀, 𝑀𝐷=𝑀𝐢, 𝐴𝐡⫽𝐢𝐷, and π‘šβˆ π‘€πΆπ·=40∘. Draw 𝑀𝐸βŠ₯𝐢𝐷 and cut 𝐢𝐷 at 𝐸, then draw 𝑀𝐹βŠ₯𝐴𝐡 and cut 𝐴𝐡 at 𝐹. Find π‘šβˆ πΆπ‘€πΈ and π‘šβˆ π΅π‘€πΉ.

Answer

We can begin by constructing the two segments 𝑀𝐸 and 𝑀𝐹, which are perpendicular to 𝐢𝐷 and 𝐴𝐡 respectively. We can also fill in the given angle information, which is π‘šβˆ π‘€πΆπ·=40∘, and mark on the diagram that 𝑀𝐷=𝑀𝐢.

In the lower triangle, △𝐢𝑀𝐷, we have two congruent line segments. This means that △𝐢𝑀𝐷 must be an isosceles triangle. There are a number of ways in which we can find π‘šβˆ πΆπ‘€πΈ. One method is to use △𝐢𝑀𝐸 and recall that the internal angle measures in a triangle sum to 180∘. Given that π‘šβˆ π‘€πΆπΈ=40∘ and π‘šβˆ πΆπΈπ‘€=90∘, we have π‘šβˆ π‘€πΆπΈ+π‘šβˆ πΆπΈπ‘€+π‘šβˆ πΆπ‘€πΈ=18040+90+π‘šβˆ πΆπ‘€πΈ=180130+π‘šβˆ πΆπ‘€πΈ=180π‘šβˆ πΆπ‘€πΈ=180βˆ’130=50.∘∘∘∘∘∘∘∘∘

Next, we need to calculate the required angle, π‘šβˆ π΅π‘€πΉ. Given that △𝐢𝑀𝐷 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 π‘šβˆ π·π‘€πΈ=π‘šβˆ πΆπ‘€πΈ=50.∘

Given that 𝐴𝐡⫽𝐢𝐷, 𝑀𝐸βŠ₯𝐢𝐷, and 𝑀𝐹βŠ₯𝐴𝐡 with 𝑀𝐸 and 𝑀𝐹 meeting at point 𝑀, then 𝐸𝐹 is a straight line. We can then observe that βˆ π·π‘€πΈ and βˆ π΅π‘€πΉ are vertically opposite and, hence, are congruent.

Therefore, we can give the answer that the unknown angle measures are π‘šβˆ πΆπ‘€πΈ=50π‘šβˆ π΅π‘€πΉ=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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