Video: Finding the Lengths of Segments in a Triangle Using the Properties of Parallel Lines and Transversals

Video Transcript

Find the values of 𝑥 and 𝑦.

What we see in the figure is the larger triangle 𝐴𝑀𝐷, where the line segments 𝐸𝐵 and 𝐹𝐶 are cutting this triangle. In addition to that, the line segments 𝐹𝐶, 𝐸𝐵, and 𝐴𝐷 are all parallel. And by the side splitter theorem, when a triangle is cut by a parallel line to one of the sides, it splits the side lengths proportionally. We see that the three segments created, 𝐸𝐷, 𝐹𝐸, and 𝑀𝐹, are all equal to each other. And since that’s the case, and since these parallel lines cut this triangle proportionally, we can say that 𝑀𝐶 is going to be equal to 𝐶𝐵, which is going to be equal to 𝐵𝐴.

If we write the statement like this, 𝑦 plus four equals four 𝑥 plus one which equals 𝑥 squared minus four, it doesn’t seem really clear what we’re solving for. So let’s break this up. If we said four 𝑥 plus one equal to 𝑥 squared minus four, we can solve for 𝑥.

Since we have 𝑥 squared, we’re dealing with a quadratic equation. And we want to set this equation equal to zero. So we subtract four 𝑥 from both sides of the equation. And then we subtract one from both sides of the equation. On the left, we’ll only have zero. And on the right, we’ll have 𝑥 squared minus four 𝑥. And then we’ll have minus four minus one, which we can combine to minus five, so that we have zero equals 𝑥 squared minus four 𝑥 minus five.

And then we can factor this quadratic. We need to find two terms that multiply together to equal negative five and add together to equal negative four, which will be positive one and negative five. Then we set both of these terms equal to zero so that we have 𝑥 plus one equals zero and 𝑥 minus five equals zero. We’ll either have 𝑥 equals negative one or 𝑥 equals positive five.

If we plug in 𝑥 equals negative one into four 𝑥 plus one and 𝑥 squared minus four, we get negative lengths, which wouldn’t work. And that means the only valid option for us is 𝑥 equals five. If 𝑥 equals five, then these set segments are equal to 21. Four times five plus one equals 21, and five squared minus four equals 21.

Now that we know that, we can create a third equation to solve for 𝑦. This is because 𝑦 plus four must also be equal to 21. If we write this, 21 equals 𝑦 plus four, and then subtract four from both sides, we see that 𝑦 must be equal to 17. Because we knew that these parallel lines cut this triangle proportionally, we were able to find that 𝑥 equals five and 𝑦 equals 17.