# Angle bisector theorem examples | Geometry | Khan Academy

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.

wonderful … good jop

Love you Sal.

Your alot better than ma piece of shit teacher. Thank You soo much ðŸ™‚

Bless your soul!!! Thank you, I have a test tomorrow and this video clarified this topic better :3

Excellent as usual!

Useful, thank you. BYU's online courses are so bad at explaining things; this really helps.

What if it's two numbers and they're both on the base?