Question Video: Using the Power of a Point Theorem for Two Secants to Find Missing Lengths Mathematics • First Year of Secondary School

Video Transcript

A circle has two secants, line segment 𝐴𝐵 and line segment 𝐴𝐷, intersecting at 𝐴. Given that 𝐴𝐸 equals three centimeters, 𝐸𝐷 equals five centimeters, and 𝐴𝐵 equals nine centimeters, find the length of 𝐵𝐶, giving your answer to the nearest tenth.

Let’s begin by adding the information we’ve been given to the diagram. First, we’re told that the length of 𝐴𝐸 is three centimeters and, secondly, that the length of 𝐸𝐷 is five centimeters. The final piece of information we’re given is that the length of 𝐴𝐵 is nine centimeters. We want to find the length of 𝐵𝐶, which is a segment of the secant 𝐴𝐵. To do this, we can recall the intersecting secants theorem, which is a special case of the power of a point theorem. This states that if 𝐴 is a point outside a circle and 𝐵, 𝐶, 𝐷, and 𝐸 are points on the circle such that the line segment 𝐴𝐵 is a secant to the circle at 𝐵 and 𝐶 and the line segment 𝐴𝐷 is a secant to the circle at 𝐷 and 𝐸, then 𝐴𝐶 multiplied by 𝐴𝐵 is equal to 𝐴𝐸 multiplied by 𝐴𝐷.

Let’s have a look at what we know. We were given that the length of 𝐴𝐵 is nine centimeters. We were also given that the length of 𝐴𝐸 is three centimeters. And we can work out the length of 𝐴𝐷 as the sum of three centimeters and five centimeters, so it’s eight centimeters. We therefore have the equation 𝐴𝐶 multiplied by nine is equal to three multiplied by eight. Now we can use this to calculate the length of 𝐴𝐶 although it isn’t 𝐴𝐶 we want to find; it’s 𝐵𝐶. But 𝐴𝐶 plus 𝐵𝐶 will give the length of the entire secant segment 𝐴𝐵, which we know to be nine. So if we can find the length of 𝐴𝐶, we can then use the second equation to determine the length of 𝐵𝐶. Simplifying the first equation, we have nine 𝐴𝐶 is equal to 24.

To solve for 𝐴𝐶, we need to divide both sides by nine. That gives 𝐴𝐶 equals 24 over nine. And dividing both the numerator and denominator by three, this simplifies to eight over three. We can then substitute this value for the length of 𝐴𝐶 into our second equation, giving eight over three plus 𝐵𝐶 equals nine. 𝐵𝐶 is then equal to nine minus eight over three. And we can perform this subtraction more simply if we express nine as 27 over three. 27 over three minus eight over three is equal to 19 over three, which, as a decimal, is equal to 6.3 recurring. The question specifies that we should give our answer to the nearest tenth. So, rounding to one decimal place, we found that the length of 𝐵𝐶 to the nearest tenth is 6.3 centimeters.