I came across this question:
In the triangle $ABC$, $\angle BAC = w$ and $\angle CBA = 2w$, where $2w$ is acute, and $BC = x$.
- Show that $AB = (3 + 4 \sin(w))x$.
- The point $D$ is the midpoint of $AB$ and the point $E$ is the foot of the perpendicular from $C$ to $AB$. Find an expression for $DE$ in terms of $x$.
- The point $F$ lies on the perpendicular bisector of $AB$ and is a distance $x$ from $C$. The points $F$ and $B$ lie on the same side of the line through $A$ and $C$. Show that the line $FC$ trisects the $\angle ACB$.
But I don't quite understand what is meant by $F$ and $B$ lie on the same side of the line via $A$ and $C$? Can someone please explain this to me?
Thank you!
As shown in this diagram, for very small $\enspace\pmb{\omega}$, $\enspace\pmb{F}\space$ can lie outside $\triangle ABC$. For larger $\enspace\pmb{\omega}\left(\le 45^o\right)$, $\space\pmb{F}\space$ lies inside $\triangle ABC$.
If you extend the perpendicular bisector "up" further from $D$ you can locate another point $F'$ on it at distance $x$ from $C$. If you extend the segment $AC$ in both directions to a line, then $F$ and $B$ are on the same side of that line while $F'$ and $B$ are on opposite sides. You have to use that fact somehow in your proof, which will fail if you try for $F'$.
Neither $F$ nor $F'$ need be inside the triangle. Clearly $F$ is not in the picture.
(The picture does not show $E$. )