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## High school geometry

### Course: High school geometry > Unit 4

Lesson 6: Proving relationships using similarity- Pythagorean theorem proof using similarity
- Exploring medial triangles
- Proof: Parallel lines divide triangle sides proportionally
- Prove theorems using similarity
- Proving slope is constant using similarity
- Proof: parallel lines have the same slope
- Proof: perpendicular lines have opposite reciprocal slopes

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# Proof: Parallel lines divide triangle sides proportionally

Prove that a line parallel to one triangle side divides other sides proportionately. Created by Sal Khan.

## Want to join the conversation?

- At2:19, why did Sal put angle 2 and angle 4 in double arcs instead of single arcs?(4 votes)
- It’s just a way of expressing the angles are congruent. If the single arcs are already being used, the double arcs are the thing you would use next - but it dosen’t really matter(9 votes)

- Around6:00in the video, I don't understand how Sal gets the EC and DB in the fractions.(4 votes)
- He starts from AE*BD=AD*EC. As he states, he divides both sides by EC first to give AE*DB/EC=AD*EC/EC, they cancel on the right, then he does the same thing by dividing by DB which cancels on the left to get the fractions.(7 votes)

- 3:20I just wanted to know why he continued the proof after saying that both triangles were similar since if they are similar then doesn't that mean that all 3 angles of 1 triangle each have a corresponding and equivalent second angle and the same thing with the sides except that they are propionate and not equivalent?(5 votes)
- If you state that N lines are of equal length, by the number of slashes you put in them. And for angles you add extra rings/slashes. (hypothetically) Is there a generally accepted way to visually state that line 'AB' of triangle ABC is proportionate to line 'XY' of triangle XYZ? It would really help me with writing notes.(3 votes)
- I don't know of any common way of noting it on a diagram, but if AB is proportional to XY, you could use a symbol used for proportionality: "∝". AB is proportional to XY would look like: AB ∝ XY.(3 votes)

- Isn't this also known as the basic proportionality theorem ?(3 votes)
- I cant make heads or toes what he means.(0 votes)
- It is simple if you watch carefully and watch the videos in order(3 votes)

- is this the thales theory?(1 vote)
- No, Thales theorem says that if you inscribe a triangle in a circle, and one if its sides is a diameter of the circle, then the angle opposite that side is right.(1 vote)

- When you insert the numbers given into the proof..One side is 1/3 while the other is 1/4 ?(1 vote)

## Video transcript

- [Instructor] We're asked to prove that if a line is parallel
to one side of a triangle, then it divides the other
two sides proportionally. So pause this video and
see if you can do that. And you might wanna leverage this diagram. All right, so let's work
through this together. So we can start with this diagram. And 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. And so 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. So really 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. So the way that we can try to do it is to establish a 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. So that tells us that these
two corresponding angles are going to be congruent. So we could say that angle one
is congruent to angle three. And the reason why, is because they are corresponding,
corresponding angles. I'm just trying to write
a little bit of shorthand. This is short for corresponding angles. That's the rationale. And we also know that angle two is congruent to angle
four for the same reason. So angle two is congruent to angle four. Once again, because they
are corresponding angles. This time we have a different transversal, corresponding angles where a transversal
intersects two parallel lines. And so now, if you look at
triangle AED and triangle ACB, you see that they have two
sets of corresponding angles that are congruent. And 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. And so we can say that triangle AED is similar to triangle ACB, ACB by angle similarity. Similarity. And then given that these two are similar, then we can set up a proportion. That tells us that the ratio
of the length of segment AE to this entire side to AC, is equal to the ratio of AD,
the length of that segment to the length of the entire thing, to AB. Now this implies, and I'm
just gonna 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, and then this is going to be equal to the length of segment AD over segment AB, its length
is the length of segment AD, AD plus segment DB, plus D B. 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. So 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. And so this is going to be equal
to the length of segment AE times AD plus DB, those segment lengths, that's gotta be equal
to length of AD times AE plus the length of segment EC. And 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. And let's see, is there anything
that I can simplify here? We'll have AE times AD on both sides. So let me just subtract AE
times AD from both sides. And so then I'm just left
with 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. And 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, length of segment AE over
the length of segment EC is equal to the length of segment AD, 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 proportionally.