Why does Prof. Loh multiply r minus d with d plus r?
-
@TSS-Graviser It's kind of neat that all of your questions about angles, arcs and Power of a Point Theorem are tying in together! Now I see that you're trying to synthesize it in your head in order to make it into a tool for yourself to use. It's very good to ask, "When exactly can I use this? When can I not use this?" If you don't mind, I'll take the long approach again and backtrack to the proof of Power of a Point and how it all ties in together.
First, let's start with the two chords that intersect in a circle.
The yellow angle inscribes the yellow arc here:
Here, with the second chord, we can connect its angle vertex in such a way that the same arc is inscribed by the second yellow angle.
Now it's time to look at the other angles of the triangles! There are two other angles that inscribe the same arc of the circle, the green ones shown here:
And we also know that the angles formed by the "X" are the same, so the two triangles drawn inside the circle are similar! (They're not necessarily congruent, because we don't know any of the lengths.)
These similar triangles are analogous to the similar triangles that we saw in the Day 7 Challenge video:
The sides \( \overline{PA}\) and \(\overline{PD}\) are matching sides that go from \( \angle \text{yellow} \) to \( \angle \text{green} \), and the sides \( \overline{PB} \) and \(\overline{PC}\) are matching sides that go from \( \angle \text{yellow} \) to the third angle (not colored). This is the reason for writing the ratios
$$ \frac{ PA}{PD} = \frac{PB}{PC} $$
If this looks confusing, perhaps try to imagine that the large triangle is twice the dimensions of the small triangle. Then \( \overline{PA}\) would be double \( \overline{PD},\) and \(\overline{PB}\) would be double \(\overline{PC}.\) Then
$$ \frac{ PA}{PD} = \frac{PB}{PC} = 2 $$
At any rate, if you multiply both sides by \( \overline{PD} \) and \( \overline{PC},\) you can cancel the denominators to get the Power of a Point theorem:
$$ PA \times PC = PD \times PB $$
What's the pattern that we can find in this formula? Well, \( \overline{PA}\) and \(\overline{PC}\) are on the same chord, \( \overline{CA},\) and \(\overline{PD}\) and \(\overline{PB}\) are on the same chord \(\overline{BD}.\) In other words, try to look for the entire chords in the picture; their segments will be the ones that you multiply together in Power of a Point. The purple line below is one chord:
And this purple line is another:
Now, here is the tool we have for applying the Power of a Point Theorem:
-
Take the two segments of one chord, and multiply them together.
-
Take the two segments of the other chord, and multiply them together.
-
These two products should equal each other.
I hope this made sense! Let me know!
-
-
Sorry I get the first part now but I don't get all those angle equations in the second half. I just want to know if angles determine side length relationships.
-
@TSS-Graviser Oh, are you referring to the equations like
$$ \frac{PA}{PD} = \frac{PB}{PC}?$$
I think I might see what you're confused about. Are you thinking that \(PA, PD, PB,\) and \(PC\) are measures of angles? They are actually lengths of lines segments.
If we were talking about the measure of angles, we would write this as
$$ \angle CPD $$
for the yellow angle, for example.
By the way, if you want to write this yourself, you can type
\\( \angle CPD \\) -
No I mean all that matching stuff and whatnot. Is there like a simplified explanation that like I can digest?
The Blade Dancer
League of Legends, Valorant: Harlem Charades (#NA1)
Discord: Change nickname if gay#7585 -
And by the way, when can power of a point be used? It seems that I can use it anywhere with intersections, but is that always the case?
-
@TSS-Graviser Oh, I see! Do you mean that you're not sure how to figure out which sides match? Try imagining that you are folding the screen like closing a book, so that the yellow angles lie on top of each other and the green angles also lie on top of each other. Do you see that \( \overline{PA}\) will lie right on top of \(\overline{PD}?\)
-
@debbie Yes! Whenever you have two chords intersecting inside a circle, you can use Power of a Point.
-
@TSS-Graviser The simplified version was found at the bottom of the earlier very long reply:
Now, here is the tool we have for applying the Power of a Point Theorem:
Take the lengths of two segments of one chord, and multiply them together.
Take the lengths of two segments of the other chord, and multiply them together.
These two products should equal each other.
I hope this made sense! Let me know!
-
Oh ok thanks. What I've been trying to say is that above in the pictures with the arcs, right, how do all these line segments on the same chord mean anything? And what's the point of labeling the orange arc and the green arc? Do they have a purpose?
-
@TSS-Graviser That's a good question; it depends on what you mean by meaningful! The line segments happen to be the side lengths of the triangles in the picture. Maybe you will see a math question that asks for one of the lengths of the triangles. In that case, knowing Power of a Point will be very meaningful!
The yellow arc goes with the yellow angle. The yellow angle is inscribed inside the yellow arc. That means that the yellow angle's measure is \( \frac{1}{2}\) of the measure of the arc.
So to answer your emailed question,
, suppose an angle opens up to exactly half of the circle. That half of the circle is an arc with measure \( 180^{\circ}.\) That means the angle is \( 90^{\circ}.\)
