Question Video: Finding Unknown Side Lengths in a Triangle Using the Angle Bisector Theorem | Nagwa Question Video: Finding Unknown Side Lengths in a Triangle Using the Angle Bisector Theorem | Nagwa

Question Video: Finding Unknown Side Lengths in a Triangle Using the Angle Bisector Theorem Mathematics • First Year of Secondary School

Given that angle 𝐴 is bisected by line segment 𝐷𝐴, 𝐴𝐵 = 38, 𝐴𝐶 = 18, and 𝐵𝐶 = 28. Determine the lengths of 𝐷𝐵 and 𝐷𝐶.

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Video Transcript

Given that angle 𝐴 is bisected by line segment 𝐷𝐴, that 𝐴𝐵 equals 38, 𝐴𝐶 equals 18, and 𝐵𝐶 equals 28, determine the lengths of 𝐷𝐵 and 𝐷𝐶.

The first thing we can do is take the information we’re given and write it down in our figure. 𝐴𝐵 equals 38. 𝐴𝐶 equals 18. We need to be careful here because we have the distance 𝐵𝐶 and 𝐵𝐶 equals 28. At this point, it might seem like there’s very little else we can say. But because we know that line segment 𝐷𝐴 bisects angle 𝐴, we can use the angle bisector theorem, which tells us that an angle bisector of a triangle divides the opposite side into two segments that are proportional to the other adjacent sides of the triangle.

We have the opposite sides 𝐷𝐶 and 𝐷𝐵 and the adjacent sides 𝐴𝐵 and 𝐴C. Because these side lengths are in proportion to one another, it means the ratio of opposite to adjacent will be equal for both sets. 𝐴𝐶 over 𝐷𝐶 must be equal to 𝐴𝐵 over 𝐷𝐵. We know that 𝐴𝐶 equals 18 and 𝐴𝐵 equals 38. But this equation on its own is not enough to solve for 𝐷𝐵 and 𝐷𝐶. So, we’ll need to write a second equation. We know that line 𝐵𝐶 equals 28 and line 𝐵𝐶 is made up of two segments: 𝐷𝐵 and 𝐷𝐶. So, we can say that 𝐷𝐵 plus 𝐷𝐶 equals 28. And now, we have two equations that we can use to solve for our missing side lengths.

We can look at our second equation and try and solve for 𝐷𝐵. That means get 𝐷𝐵 by itself. To do that, we would subtract 𝐷𝐶 from both sides of the equation. And then, we could say that 𝐷𝐵 equals 28 minus 𝐷𝐶. Now that we know what 𝐷𝐵 is equal to, we can take this information and plug it back in to our first equation. If 18 over 𝐷𝐶 equals 38 over 𝐷𝐵 and 𝐷𝐵 equals 28 minus 𝐷𝐶, then we plug in 28 minus 𝐷𝐶 for the denominator here. But we now have two 𝐷𝐶 variables in the denominator. And we’d like to move these to the numerator. To do that, we cross multiply the numerators and the denominators. 18 times 28 minus 𝐷𝐶 is equal to 38 times 𝐷𝐶. On the left, we’ll need to distribute this times 18. 18 times 28 equals 504 and 18 times negative 𝐷𝐶 equals negative 18 𝐷𝐶.

Now, we have 504 minus 18 𝐷𝐶 equals 38 𝐷𝐶. We want to get 𝐷𝐶 on the same side of the equation. And so, we add 18 𝐷𝐶 to both sides. 38 plus 18 equals 56. So we have 504 equals 56 𝐷𝐶. And then, we divide both sides of the equation by 56 504 divided by 56 equals nine and 56 𝐷𝐶 divided by 56 equals 𝐷𝐶. And that means 𝐷𝐶 equals nine. We can go ahead and add this to our diagram. 𝐷𝐶 equals nine. That’s the same thing as 𝐶𝐷 equals nine. It doesn’t matter which order you say the endpoints. Because we know that 𝐷𝐶 equals nine, we can say that 𝐷𝐵 equals 28 minus nine. 28 minus nine is 19. So, we can say that 𝐷𝐵 equals 19.

Let’s perform a few checks here. First, does nine plus 19 equal 28? It does. And secondly, let’s check these proportional relationships. Is 18 over nine equal to 38 over 19? If we divide 18 by nine, we get two. And if we divide 38 by 19, we also get two. Two is equal to two. And so, these relationships are in proportion with each other. And we can say that 𝐷𝐵 equals 19 and 𝐷𝐶 equals nine.

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