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Using the angle bisector theorem

Sal uses the angle bisector theorem to solve for sides of a triangle. Created by Sal Khan.

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Video transcript

I thought I would do a few examples using the angle bisector theorem. So in this first triangle right over here, we're given that this side has length 3, this side has length 6. And this little dotted line here, this is clearly the angle bisector, because they're telling us that this angle is congruent to that angle right over there. And then they tell us that the length of just this part of this side right over here is 2. So from here to here is 2. And that this length is x. So let's figure out what x is. So the angle bisector theorem tells us that the ratio of 3 to 2 is going to be equal to 6 to x. And then we can just solve for x. So 3 to 2 is going to be equal to 6 to x. And then once again, you could just cross multiply, or you could multiply both sides by 2 and x. That kind of gives you the same result. If you cross multiply, you get 3x is equal to 2 times 6 is 12. x is equal to, divide both sides by 3, x is equal to 4. So in this case, x is equal to 4. And this is kind of interesting, because we just realized now that this side, this entire side right over here, is going to be equal to 6. So even though it doesn't look that way based on how it's drawn, this is actually an isosceles triangle that has a 6 and a 6, and then the base right over here is 3. It's kind of interesting. Over here we're given that this length is 5, this length is 7, this entire side is 10. And then we have this angle bisector right over there. And we need to figure out just this part of the triangle, between this point, if we call this point A, and this point right over here. We need to find the length of AB right over here. So once again, angle bisector theorem, the ratio of 5 to this, let me do this in a new color, the ratio of 5 to x is going to be equal to the ratio of 7 to this distance right over here. And what is that distance? Well, if the whole thing is 10, and this is x, then this distance right over here is going to be 10 minus x. So the ratio of 5 to x is equal to 7 over 10 minus x. And we can cross multiply 5 times 10 minus x is 50 minus 5x. And then x times 7 is equal to 7x. Add 5x to both sides of this equation, you get 50 is equal to 12x. We can divide both sides by 12, and we get 50 over 12 is equal to x. And we can reduce this. Let's see if you divide the numerator and denominator by 2, you get this is the same thing as 25 over 6, which is the same thing, if we want to write it as a mixed number, as 4, 24 over 6 is 4, and then you have 1/6 left over. 4 and 1/6. So this length right over here is going, oh sorry, this length right over here, x is 4 and 1/6. And then this length over here is going to be 10 minus 4 and 1/6. What is that? 5 and 5/6.