Hello, MATNTRNG!
$\displaystyle \text{Use the angles sum of a triangle theorem and the linear pair theorem}$
$\displaystyle \text{to prove the following result:}$
$\displaystyle \text{If }\Delta ABC\text{ is a right triangle with }C = 90^o\text{ and }O\text{ is the midpoint of }AB,$
$\displaystyle \text{then: }\:\angle BOC \:=\: 2\!\cdot\!\angle BAC $
I have my own proof of this . . . Code:
A o
* * r
* @ *
* * O
* o
* * 2@* r
* @ * r *
* * *
C o * * * o B
Right triangle $\displaystyle ABC$ can be inscribed in a semicircle.
. . $\displaystyle O$ is the center of the circle
. . and $\displaystyle OA, OB, OC$ are radii $\displaystyle \,r$ of the circle.
Hence, $\displaystyle \Delta AOC$ is isosceles and $\displaystyle \angle OAC = \angle ACO = \theta$
$\displaystyle \angle BOC$ is an exterior angle of $\displaystyle \Delta AOC$.
. . Hence: .$\displaystyle \angle BOC \:=\:\angle OAC + \angle ACO \;=\;\theta + \theta \;=\;2\theta$
Therefore: .$\displaystyle \angle BOC \;=\;2\!\cdot\!\angle OAC \;=\;2\!\cdot\!\angle BAC$