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

# Lesson Video: Corollaries of Isosceles Triangle Theorems Mathematics

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

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### Video Transcript

In this video, we will learn how to use the corollaries of the isosceles triangle theorems to find missing lengths and angles in isosceles triangles. We can recall that an isosceles triangle is simply a triangle that has two congruent sides. And when we are talking about isosceles triangles, we use two important terms. They are the legs and the base. The congruent sides are called the legs of the triangle and the third side is the base.

Now, because isosceles triangles have two congruent sides, this leads us to an important angle property of isosceles triangles. It is the isosceles triangle theorem, and it states that if two sides of a triangle are congruent, then the angles opposite those sides are congruent. So knowing that the two sides are congruent means that in fact the two base angles are congruent. The remaining angle in the isosceles triangle is referred to as the vertex angle. So we know that isosceles triangles by definition have two congruent sides. And by this theorem, it means that they also have two congruent angles. And the converse of this theorem is also true. That is, if two angles of a triangle are congruent, then the sides opposite those angles are also congruent.

We will now consider a number of corollaries to these theorems. These corollaries will allow us to identify additional geometric properties about isosceles triangles. Let’s see the first of these corollaries. This corollary states that the median of an isosceles triangle from the vertex angle bisects it and is perpendicular to the base. We can prove this corollary in the following way. Let’s take this isosceles triangle 𝐴𝐵𝐶, and we draw the median from 𝐴 to create the point 𝐷. The median of a triangle is a line segment joining a vertex to the midpoint of the opposite side and therefore bisecting that side. We can therefore say that 𝐶𝐷 is congruent to 𝐵𝐷. We can also write that 𝐴𝐶 is equal to 𝐴𝐵 because we know that the triangle is isosceles and the line segment 𝐴𝐷 is a shared side in the two triangles 𝐴𝐷𝐶 and 𝐴𝐷𝐵.

And as there are now three congruent pairs of sides, then we can say that triangle 𝐴𝐷𝐶 is congruent to triangle 𝐴𝐷𝐵 by the SSS, or side-side-side, congurrency criterion. That means that the measure of angle 𝐴𝐷𝐶 is congruent to the measure of angle 𝐴𝐷𝐵. Furthermore, since the line segment 𝐵𝐶 is a straight line, then both of these angles 𝐴𝐷𝐵 and 𝐴𝐷𝐶 must be 90 degrees. So we have proved this corollary. The median of an isosceles triangle from the vertex angle bisects it and is perpendicular to the base. Because of this corollary, we can also notice that a useful property of the median of an isosceles triangle is that it forms the axis of symmetry of this triangle because it splits the isosceles triangle into two congruent right triangles.

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

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

In the figure, we can observe that triangle 𝐴𝐵𝐶 appears to be an isosceles triangle. Although we can’t prove this, we can use some of the properties of isosceles triangles to help. We can recall that an isosceles triangle is a triangle that has two congruent sides and the median of an isosceles triangle from the vertex angle is a perpendicular bisector of the base. And in this question, we are interested in the perpendicular bisector of the line segment 𝐵𝐶, which would be the base of this triangle. That means that the line segment 𝐴𝐷 is only a perpendicular bisector of line segment 𝐵𝐶 in the case of an isosceles triangle. So when this is an isosceles triangle, the two legs 𝐴𝐵 and 𝐴𝐶 will be congruent. And if line segment 𝐵𝐶 is bisected, then 𝐵𝐷 is equal to 𝐶𝐷.

So let’s take this first equation 𝐴𝐵 is equal to 𝐴𝐶 and fill in the given expressions for these lengths. This would give us three 𝑥 plus two is equal to five 𝑦 plus three. We can rearrange this by subtracting three from both sides to give us three 𝑥 minus one is equal to five 𝑦. We can’t solve this just yet as there are two unknowns, so let’s label this with equation one. For the second equation, we can fill in the lengths for 𝐵𝐷 and 𝐶𝐷. This gives five 𝑦 minus one is equal to 10 minus three 𝑥. Adding one to both sides gives us a second equation of five 𝑦 is equal to 11 minus three 𝑥. We can then find the values of 𝑥 and 𝑦 by solving these two equations simultaneously either by substitution or elimination.

We might notice however that both equations have a term of five 𝑦 on one side of the equation. We can therefore write that three 𝑥 minus one is equal to 11 minus three 𝑥. Then, by adding three 𝑥 and one to both sides of the equation either in one step or two steps, we have that six 𝑥 is equal to 12, so 𝑥 is equal to two. We can then substitute 𝑥 is equal to two into either equation one or two. So using equation one, we have that three times two minus one is equal to five 𝑦. That gives us that five is equal to five 𝑦, and so 𝑦 is equal to one. We can therefore give the answer that the line segment 𝐴𝐷 is a perpendicular bisector of the line segment 𝐵𝐶 when 𝑥 is equal to two and 𝑦 is equal to one.

We’ll now see another corollary. This corollary tells us that the bisector of the vertex angle of an isosceles triangle is a perpendicular bisector of the base. Let’s look at what we mean by this corollary by taking the triangle 𝐸𝐹𝐺, which is isosceles. We can draw in the bisector of the vertex angle, which will meet the line segment 𝐺𝐹 at the point 𝐻. And as this is a bisector, we can say that the measure of angle 𝐺𝐸𝐻 is equal to the measure of angle 𝐹𝐸𝐻. This corollary would then allow us to state that the base has been bisected, so 𝐺𝐻 is equal to 𝐹𝐻. And because it does this at right angles, then the measure of angle 𝐸𝐻𝐺 and the measure of angle 𝐸𝐻𝐹 are both 90 degrees.

We’ll now see how we can apply this in the following example.

Fill in the blank. In this figure, if 𝐴𝐵 equals 𝐴𝐶, the intersection of line segments 𝐴𝐷 and 𝐵𝐶 is equal to the set containing 𝐷, where 𝐷𝐶 equals eight centimeters and the measure of angle 𝐶𝐴𝐷 equals the measure of angle 𝐵𝐴𝐷 equals 35 degrees, the length of line segment 𝐵𝐷 is what centimeters. The second part of this question asks us to find the measure of angle 𝐴𝐶𝐵.

The first very important piece of information we are given is that 𝐴𝐵 is equal to 𝐴𝐶. As this triangle 𝐴𝐵𝐶 has two congruent sides, then it means that triangle 𝐴𝐵𝐶 is an isosceles triangle. We are also told that this line segment 𝐴𝐷 intersects the base 𝐵𝐶 at the point 𝐷. We can also fill in the angle measures that the measure of angle 𝐶𝐴𝐵 and 𝐵𝐴𝐷 are both 35 degrees. We can observe that since these two angles are equal in measure, it means that the vertex angle 𝐴 of this isosceles triangle has been bisected.

We can recall that one of the corollaries to the isosceles triangle theorems states that the bisector of the vertex angle of an isosceles triangle is a perpendicular bisector of the base. So in triangle 𝐴𝐵𝐶, the base 𝐵𝐶 must be bisected. And that means that the line segment 𝐵𝐷 is congruent to the line segment 𝐷𝐶. And therefore, 𝐵𝐷 must also be eight centimeters. Therefore, the value of eight would complete the blank in the first part of this question.

We can now look at the second part of the question where we need to determine the measure of angle 𝐴𝐶𝐵. To do this, let’s go back to the corollary where we are told that this bisector of the vertex angle is a perpendicular bisector of the base. This means that the measure of angle 𝐴𝐷𝐶 is 90 degrees. By considering triangle 𝐴𝐶𝐷 and using the fact that the interior angles in a triangle sum to 180 degrees, we can work out that the measure of angle 𝐴𝐶𝐷, which is the same as the measure of angle 𝐴𝐶𝐵, must be equal to 180 degrees subtract 35 degrees plus 90 degrees. This simplifies to 55 degrees. We have therefore answered the second part of this question, but of course it’s always worthwhile checking our answers.

For the second part of this question, we recall that we have already said that triangle 𝐴𝐵𝐶 is an isosceles triangle. That means that if angle 𝐴𝐶𝐵 is 55 degrees, then the other base angle of this isosceles triangle 𝐴𝐵𝐶 must also be 55 degrees. Adding the three interior angle measures would give us 55 degrees plus 55 degrees plus 70 degrees. And that would give us 180 degrees, which we know is the sum of the interior angles in the triangle. And therefore, we have confirmed that the measure of angle 𝐴𝐶𝐵 must be 55 degrees.

We will now see the final corollary to the isosceles triangle theorems. This corollary 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. Let’s look at this diagrammatically with the isosceles triangle 𝑃𝑄𝑅. If we have a straight line through the vertex angle perpendicular to the base, then the base is bisected and the vertex angle is bisected. And we’ll now see how we can apply this corollary in the next example.

Find the measure of angle 𝐷𝐴𝐵.

In this triangle 𝐴𝐵𝐶, we can observe that we have two congruent sides marked. 𝐴𝐵 is equal to 𝐴𝐶. We can therefore recognize that triangle 𝐴𝐵𝐶 must be an isosceles triangle. We can see on the diagram that the measure of angle 𝐴𝐷𝐵 is 90 degrees. In other words, we can say that the line segment 𝐴𝐷 is perpendicular to the base 𝐶𝐵. And this is a very important piece of information because it means that we can apply one of the corollaries to the isosceles triangle theorems. This corollary 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.

Since we know that 𝐴𝐷 is perpendicular to the base, then the important thing is that this line segment 𝐴𝐷 bisects the vertex angle. The angle 𝐷𝐴𝐵 that we need to work out the measure of is part of the vertex angle of the isosceles triangle. Because we know that this angle 𝐶𝐴𝐵 has been bisected, then the measure of angle 𝐶𝐴𝐷 is congruent to the measure of angle 𝐷𝐴𝐵. Both of these angles will be 25 degrees. We can therefore give the answer that the measure of angle 𝐷𝐴𝐵 is 25 degrees.

We will now summarize the key points of this video. We began by recalling the definition of an isosceles triangle, which is that it is a triangle that has two congruent sides. The isosceles triangle theorem tells us that if two sides of a triangle are congruent, then the angles opposite those sides are also congruent. The converse of this theorem states that if two angles of a triangle are congruent, then the sides opposite those angles are also congruent.

We then saw three separate corollaries, the first of which states that the median of an isosceles triangle from the vertex angle is a perpendicular bisector of the base and bisects the vertex angle. The second corollary states that the bisector of the vertex angle of an isosceles triangle is a perpendicular bisector of the base. The third corollary 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. Finally, we also noted that the axis of symmetry of an isosceles triangle is the median that bisects the base.