圆幂定理

Let's say that we are given the above diagram, and we want to find the length of $AE$. Don't worry, we are told the lengths of $BE$, $DE$, and $CE$, so if we can find $AE$ in terms of those lengths, life is good. Unfortunately, $A$, $B$, $C$, and $D$ are just four random points on a random circle, and we definitely can't assume that $E$ is the center of the circle.

If we just connect $A$ with $B$ and $C$ with $D$,

Boom! Inscribed angles, triangles, you-name-it. It's incredible what adding a few lines can do.

The angles $\angle BAC$ and $\angle BDC$ intersect the same arc, so they are equal! Similarly, the angles $\angle DBA$ and $\angle DCA$ intersect the same arc, so they are also equal. And of course, the vertical angles $\angle AEB$ and $\angle DEC$ are equal. Any two of these three observations would tell us that we've got two triangles with equal angles. Wow, triangles with equal angles.. Looks suspicious.

"WE ARE SIMILAR!" the triangles shout. Similarity tells us that the sides of these two triangles are proportional to each other. Thus, we know $$\frac{AE}{BE} = \frac{DE}{CE}$$ $$AE \cdot CE = BE \cdot DE$$ $$AE =\frac{ BE \cdot DE }{CE}.$$

Awesome, we found $AE$! That's great. However, this wasn't actually just a random situation. The common value of the products in the equation $AE \cdot CE = BE \cdot DE$ depends only on the powerful point $E$ and is therefore called the power of the point $E$ with respect to this circle. Also, the relationship $AE \cdot CE = BE \cdot DE$ is so important that it is called the Power of a Point Theorem. Turns out this theorem holds in a variety of circle and point configurations! If you are curious, you can verify that the Power of a Point Theorem holds in the two set-ups below. Use steps very similar to those we just used in your solutions.

This post addresses the tangent/degenerate case of power of a point.

Power of a Point states that, in the following diagram,

$$\overline{PB} \times \overline{AP}= \overline{PC} \times \overline{PD}$$ always holds. But what happens if $\overline{CP}$ or $\overline{BP}$ becomes, instead of a secant, a tangent to the circle? i.e.

so $A$ coincides with $B$?

Amazingly (or un-amazingly), Power of a Point still holds, just in a different form, namely $$\overline{BP}^2 = \overline{PC} \times \overline{PD}$$, and can be proved with the same method used to prove the "original" Power of a Point.

Essentially, the two similar triangles are $BDP$ and $BCP$ because $\angle DBP= \angle BCP$ by inscribed angles and $\angle BPD = \angle BPC$. From this, we get $$\begin{align*} \frac{\overline{BP}}{\overline{PC}} &= \frac{\overline{PD}}{\overline{BP}} \\ \overline{BP}^2 &= \overline{PC}\times\overline{PD} \end{align*}$$ and we are done.

When two chords intersect, as seen above, they are split in such a way where the product of the chord segments are equal.

In Diagram $DE(EB) = AE(EC)$