Angle bisector theorem examples | Geometry | Khan Academy

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.

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