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Show that a line parallel to the base of a triangle intersecting the other two sides, forms a traingle similar to the given triangle.

54 in

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 9

Elements of Geometry

Derivatives

Campbell University

Harvey Mudd College

Baylor University

Idaho State University

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

04:37

Prove that a line that div…

00:23

Prove: If a line bisects b…

01:48

Show that the line through…

for this problem. We have a triangle, and we're told that we have a line parallel to it to the base of this triangle that intersects the other two sides so parallel to the base, intersecting the other two sides. So that's what we have. And we want to show that the small triangle at the top there is similar to the large triangle. So in other words, let me just label some of these points here A, B, C, D and E. So we want to show the triangle D B E. The small triangle is similar to triangle A B C. The large triangle. Well, how can we do that? Well, if we have similar triangles, there are two different things to different ways that makes little bit easy to show if they're similar or not. First, if I know the sides and I can show that the sides have the same proportions between one and the other, that would be similar triangles. I know nothing about the sides of any of these, uh, other than the two are parallel. I know nothing about the size is how long they are. That's not going to work very well. another way we can show that two triangles are similar is if we show that all of their angles are equal. I mean not equal to each other, but equal from one to the other. So in other words, angle D an angle a are equal angle be well, this case B and B are equal. We've already shown that one, but also that angle E and C are equal. If we can show that each corresponding pair of angles are equal, we know that the two triangles are similar. So let's see if we can use this. I'm going to leave the little check marks over to the B because we've shown that those are equal because they're the same angle. They're both up here. They will share that top angle. But what about the other pairs? Well, let's look at A and D first. We have parallel lines, the line that connects a C. And I'm just gonna It just kind of got this out. We have that line there. The line connecting A and C and the line connecting D and E are parallel. So our line A B is a trans versatile. It crosses both of those parallel lines. Well, what do we know about the angles created by a transfer cell? I know that corresponding angles are the same. So this angle corresponds to that angle because they are the other corresponding along that Trans Verceles. Those are angles, DNA. They are indeed the same. Now, if I look at e N. C, I can make the same argument on this side B. C. That side of my triangle is a trans versatile across these parallel lines, so corresponding angles are equal, which means E and C are equal. All of my angles for one triangle. Exactly. Match the other one. So, yes, these are indeed similar.

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