Hi
Mark Powell, welcome to
MHB!:)
For the first problem, the only formula that you required to use to achieve to the result is the Pythagoras' theorem, where it states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.
View attachment 2610
First, join $BG$ and since $CB$ is a tangent at $B$, we know $AB$ and $CB$ are perpendicular. Thus, we have one right triangle $ABC$.
Next, we're told that $AB$ is the diameter of the circle centered at $O$, so, $\angle AGB=90^{\circ}$. We can tell also $\angle BGC=90^{\circ}$ by the supplementary angle theorem. Therefore both $BGA$ and $BCG$ are right triangles.
[TABLE="class: grid, width: 1000"]
[TR]
[TD]Triangle $AGB$:[/TD]
[TD]Triangle $BGC$:[/TD]
[TD]Triangle $ABC$:[/TD]
[/TR]
[TR]
[TD]Now, we apply the Pythagoras' theorem to the right triangle $AGB$ and note that $AB=2\text{radius}=2(6)=12$, we have:[/TD]
[TD]Again, reapply the Pythagoras' theorem to the right triangle $BGC$, we have:[/TD]
[TD]So Pythagoras' theorem tells us for the third time that:[/TD]
[/TR]
[TR]
[TD]$AG^2+BG^2=AB^2$
$8^2+BG^2=12^2$ this gives
$BG^2=80$[/TD]
[TD]$BG^2+CG^2=BC^2$
Remember that our aim is to find the value for $BC$ and we don't have the value for $CG$ yet, so we need to look for it by considering the right triangle $ABC$.[/TD]
[TD]$AB^2+BC^2=AC^2$
$12^2+BC^2=(AG+CG)^2$ $(AC=AG+CG)$
$12^2+BC^2=(8+CG)^2$ $(AG=8)$
By expanding the equation, we get:
$12^2+BC^2=64+18CG+CG^2$
$12^2+(BC^2-CG^2)=64+18CG$(*)[/TD]
[/TR]
[/TABLE]Up to this point, we need to be aware of what we have already found, e.g. from $BG^2+CG^2=BC^2$ and $BG^2=80$, the difference for $BC^2$ and $CG^2$ is $BC^2-CG^2=BG^2=80$
If we replace it into the equation (*), we get:
$12^2+80=64+18CG$
Solve this equation for $CG$ we see that
$\dfrac{12^2+80-64}{18}=CG$
$CG=10$
The last step would be to substitute the values for $CG=10$ and $BG^2=80$ into the equation $BG^2+CG^2=BC^2$, we find that $BC^2=80+10^2=180=36(5)$, therefore $BC=6\sqrt{5}$.
