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Using the angle bisector theorem

CCSS.Math:

Video transcript

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 three this side has length six and then 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 two so from here to here is two 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 three to two is going to be equal to six to X and we can just solve for x so three to two is going to be equal to six to X and then once again you can you could just cross multiply or you could multiply both sides by 2 and X it kind of gives you the same result if you cross multiply you get 3x 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 realize 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 it has a 6 and a 6 and then the base right over here is 3 it's kind of 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 you know if we call this point a and this point right over here if we want we need to find the length of a B 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 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 10 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 the denominator by 2 you get this is the same thing as 25 over 6 which is the same thing if we want to write 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 to 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 so what is that 5 & 5 6