Why consider (x–u)(x–v) = 0 in Thales' theorem converse proof

[murshid_islam]

Homework Statement

Proof of the converse of Thales' theorem

Relevant Equations

(x–u)(x–v) = 0

I was watching this series of videos of Eddie Woo proving Thales' theorem and its converse. I didn't understand this part (at 2:15) where he considered (x–u)(x–v) = 0. He later used the result he got from considering that. But why consider it in the first place?

 

[anuttarasammyak]

$$(x-x_1)(x-x_2)=(x-\frac{x_1+x_2}{2})^2-(\frac{x_1+x_2}{2})^2+x_1x_2=(x-\frac{x_1+x_2}{2})^2-(\frac{x_1-x_2}{2})^2$$ or
$$(x-x_1)(x-x_2)=(x-\frac{x_1+x_2}{2}-\frac{x_1-x_2}{2})(x-\frac{x_1+x_2}{2}+\frac{x_1-x_2}{2})=(x-\frac{x_1+x_2}{2})^2-(\frac{x_1-x_2}{2})^2$$
Apply the same method to y. That should be sufficient. You may ignore his '=0' if it causes any confusion.