A circle has two chords, the line segments 𝐴𝐶 and 𝐵𝐷, intersecting at 𝐸. Given that the ratio 𝐴𝐸 to 𝐵𝐸 equals one to three and 𝐶𝐸 equals six centimeters, find the length of 𝐷𝐸.
Let’s add the information we’ve been given to the diagram first. We’re told that the length of 𝐶𝐸 is six centimeters. And we’re also told that the ratio of the length of the line segments 𝐴𝐸 to 𝐵𝐸 is one to three. We can therefore express the lengths of these two line segments as 𝑥 centimeters and three 𝑥 centimeters for some nonzero value of 𝑥. We’re asked to find the length of the line segment 𝐷𝐸. And we see that the information we’ve got concerns the lengths of line segments in two different chords of a circle.
We can therefore recall a basic case of the power of a point theorem, which relates the lengths of line segments in two different chords. Let 𝐸 be a point inside circle 𝑀. If 𝐴, 𝐵, 𝐶, and 𝐷 are points on the circle such that the line segments 𝐴𝐶 and 𝐵𝐷 are two intersecting chords at 𝐸, then 𝐴𝐸 multiplied by 𝐶𝐸 is equal to 𝐵𝐸 multiplied by 𝐷𝐸. This is exactly the setup we have here. We know the length of 𝐶𝐸. It’s six centimeters. And we have expressions for the lengths of 𝐴𝐸 and 𝐵𝐸. It’s the length of 𝐷𝐸 that we want to find. So we can substitute the values or expressions we know and form an equation. 𝑥 multiplied by six is equal to three 𝑥 multiplied by 𝐷𝐸.
To find the value of 𝐷𝐸, we need to divide both sides of this equation by three 𝑥. And remember, we said 𝑥 was nonzero, so it’s fine to do this. We have six 𝑥 over three 𝑥 is equal to 𝐷𝐸. And again, as 𝑥 is nonzero, we can cancel a factor of 𝑥 in the numerator and denominator. This leaves six over three is equal to 𝐷𝐸. And of course six divided by three is equal to two. So by recalling the intersecting chords theorem, which is a special case of the power of a point theorem, we found that the length of 𝐷𝐸 is two centimeters.