Proof: The product of segments of intersecting secants
About this video
The video proves that for two secant segments drawn from an outside point A, AE x AD = AC x AB (whole secant times its external part is the same for both). It draws segments BE and DC, shows triangles ACD and AEB are similar by angle-angle (inscribed angles on the same arc plus the shared angle A), and cross-multiplies the ratio of corresponding sides.
You'll be able to
- state the secant-secant product rule: whole x external part is equal for both secants.
- explain why triangles ACD and AEB are similar.
- use the rule to find a missing secant length.
Key words
- secant
- A line that cuts through a circle at two points.
- secant segment
- A segment from a point outside the circle that passes through the circle to its far side.
- external part
- The piece of the secant segment between the outside point and the circle.
- inscribed angle
- An angle with its vertex on the circle; it is half the arc it intercepts.
- similar triangles
- Triangles with equal angles, so their sides are in proportion.
Transcript
- [Instructor] What we wanna do in this video is prove that if we have these two secant segments, AE and AC, and of course, they both start from A, over here, if we take the product, let's just start with segment, secant segment AE. If we take that length and take the product of the part of the secant segment that is external to the circle AD, so AE, that length, times AD, that should be equal to the product of the same segments on that other secant segment.
So the whole thing, AC, times the part that is external, times AB. So let's prove this and like we've done in many videos, we will do this by constructing some triangles. So let's set up a signal, a segment, I should say, BE and then another one, DC. And what I wanna show you now is that triangle ACD, ACD is, well, I'm gonna show you it's similar, but I won't write it down just yet, but we're gonna show that it's similar to triangle AEB, AEB.
How do we show it's similar? Well, we can see that this angle is going to be half the measure of this yellow arc right over here, 'cause if we have an inscribed angle, it is half the measure of the arc that it intersects. And this one also intersects the same arc and it's also an inscribed angle.
So we know that measure of angle ACD, and measure of angle AEB are the same. We also know that angle A is in both of these triangles. So we have two congruent angles in both of these triangles. So we know that these two are similar because of angle angle similarity.
Similarity. Similarity is what gives us this conclusion. Well, that's really useful 'cause now we can set up the ratio between corresponding sides of these two similar triangles. So we know that the ratio of say AD, the ratio of the length of AD, to the length of AC, so that's the two, the short and the long side, that's not blue here.
So AD to AC, AC, should be equal to the ratio of AB to AE, AB to AE. And now, we can multiply both sides by both denominators, by both AC and AE, both those lengths. So AC and AE, this cancels with that, that cancels with that. And we are left with length of segment AE times length of segment AD is equal to length of segment AB times length of segment AC which is exactly what we set out to prove.
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