Proof: Parallel lines divide triangle sides proportionally | Similarity | Geometry | Khan Academy

We're asked to prove that if a line is parallel to one side of a triangle, then it divides the other two sides proportionately. So pause this video and see if you can do that, and you might want to leverage this diagram.

Alright, so let's work through this together. We can start with this diagram. What we know is that segment ED is parallel to segment CB. So we can write that down: segment ED is parallel to segment CB. Segment ED is what they're talking about; that is a line or a line segment that is parallel to one side of the triangle.

Given what we know and what's already been written over here on this triangle, we need to prove another way of writing it. Another way of saying it divides the other two sides proportionately is that the ratio between the part of the original triangle side that is on one side of the dividing line to the length on the other side is going to be the same on both sides that it is intersecting.

So another way to say that it divides the other two sides proportionately, if we look at this triangle over here, it would mean that the length of segment AE over the length of segment EC is going to be equal to the length of segment AD over the length of segment DB. This statement right over here, and what I underlined up here, are equivalent given this triangle.

The way that we can try to do it is to establish similarity between triangle AED and triangle ACB. So how do we do that? Well, because these two lines are parallel, we can view segment AC as a transversal intersecting two parallel lines. That tells us that these two corresponding angles are going to be congruent. So we could say that angle 1 is congruent to angle 3, and the reason why is because they are corresponding.

I'm just trying to write a little bit of shorthand; this is short for corresponding angles—that's the rationale. We also know that angle 2 is congruent to angle 4 for the same reason. So angle 2 is congruent to angle 4 once again because they are corresponding angles, this time we have a different transversal, corresponding angles where the transversal intersects two parallel lines.

Now, if you look at triangle AED and triangle ACB, you see that they have two sets of corresponding angles that are congruent. If you have two sets of corresponding angles, that means that all of the angles are congruent, and you actually see that over here if you care about it. But two is enough, but you actually have a third because angle, I guess you call it BAC, is common to both triangles.

So we can say that triangle AED is similar to triangle ACB by angle-angle similarity. Then given that these two are similar, we can set up a proportion that tells us that the ratio of the length of segment AE to this entire side AC is equal to the ratio of AD, the length of that segment, to the length of the entire thing AB.

Now this implies that I'm just going to start writing it to the right here to save space; this is the same thing as the ratio of AE over AC is AE plus EC's length. So AE's length plus EC's length. This is going to be equal to the length of segment AD over segment AB's length, which is the length of segment AD plus segment DB.

Now, really what I need to do is figure out how do I algebraically manipulate it so I get what I have up here. Let me scroll down a little bit. One way I could try to simplify this is to essentially cross multiply; that's equivalent to multiplying both sides by both of these denominators, and we've covered that in other videos.

So this is going to be equal to the length of segment AE times AD plus DB—those segment lengths. That's got to be equal to the length of AD times AE plus the length of segment EC. I can distribute this over here: I have length of segment AE times length of segment AD plus length of segment AE times length of segment DB is equal to length of segment AD times length of segment AE plus length of segment AD times length of segment EC.

Let's see, is there anything that I can simplify here? Well, I have an AE times AD on both sides, so let me just subtract AE times AD from both sides. Then I'm just left with that this is equal to that.

So, scroll down a little bit more and let me actually just rewrite this cleanly. So I have AE times DB is equal to AD times EC. These are all the segment lengths right over here. Now, if you divide both sides by EC, you're going to get an EC down here, and then this would cancel out. Then if you divide both sides by DB, this will cancel out and you'll get a DB right over here.

So if you just algebraically manipulate what we just had over there, you get that the length of segment AE over the length of segment EC is equal to the length of segment AD over the length of segment DB, which is exactly what we wanted to prove. That this line right over here that is parallel to this side over here divides the other two sides proportionately.