Hello, thereddevils!
The first part has theorem . . .
Vertices $\displaystyle B$ and $\displaystyle C$ of $\displaystyle \Delta ABC$ lie on the circumference of a circle.
$\displaystyle AB$ and $\displaystyle AC$ cut the circumference of the circle at $\displaystyle X$ and $\displaystyle Y$ respectively.
Show that: .$\displaystyle \angle CBX + \angle CYX \:=\:180^o$ Theorem:
. . If a quadrilateral is inscribed in a circle, the opposite angles are supplementary.
Code:
X * * *
o *
* *
* o Y
* *
* *
* o C
o *
B * *
* *
* * *
Draw chords $\displaystyle XY,\: YC,\: CB,\: BX.$
. . $\displaystyle \begin{array}{cccccccc}\angle Y \:=\:\frac{1}{2}\text{arc}(XBC) & [1] \\ \\[-4mm] \angle B \:=\:\frac{1}{2}\text{arc}(XYC) & [2] \end{array}$
Add [1] and [2]: .$\displaystyle \angle Y + \angle B \;=\;\tfrac{1}{2}\text{arc}(XBC) + \tfrac{1}{2}\text{arc}(XYC)$
. . . . . . . . . . . . $\displaystyle \angle Y + \angle B \;=\; \tfrac{1}{2}\bigg[\text{arc}(XBC) + \text{arc}(XYC)\bigg] \;=\;\tfrac{1}{2}(360^o)$
. . . . . . . . . . . . $\displaystyle \angle Y + \angle B \;=\;180^o$