application of power of a point and ptolemy's theorem Let ABC be a triangle with side lengths $AB = 24, CA = 34, BC = 50$. Call the circle with diameter $BC$ ω with O as the center. Let $BA$ intersect $ω$ at $D$ ($D \ne B$) and $CA$ intersect ω at $E$ ($E\ne C$). Let the reflection of $D$ over $O $be defined as $D'$, and have $OA$ and $D'E$ intersect at $F$. Let$ G$ be a point on segment $BC$ such that $CD = 7BG$. Find the length of segment $FG$.
So I have tried to use power of a point with Ptolemy's theorem but doesn't find it works well since there aren't really enough info... Any help on how to do this will be greatly appreciated
EDIT: apparently this problem is really complex (all long geometry problem is complex when there isn't similar triangle! jk) So the simplest way is to use law of sine+cosine, property of cyclic quadrilaterals? Is there another solution to this with less intuition/ calculation? I just think I could never think about adding of point f' which the solution did :(
 A: Here's a scheme of the solution.


*

*By the cosine rule you can find angles $\angle ABC$ and $\angle ACB$:
$$
\cos(\angle ABC)={4\over5},\quad \sin(\angle ABC)={3\over5},\quad
\cos(\angle ACB)={154\over170},\quad \sin(\angle ACB)={72\over170}.
$$

*From that you get $CD=BD'=30$.

*Let $F'$ be the point where line $ED'$ meets the tangent at $B$, and apply the sine rule to triangle $BD'F'$. By taking into account that 
$\angle BD'F'=\angle BCA$ and $\angle F'BD'=\pi/2+\angle DCB=\pi-\angle ABC$, one gets: $\displaystyle BF'={1800\over29}$.

*By the cosine rule applied to $BAO$ you can first compute $OA=\sqrt{241}$ and then find $\cos\angle BOA$, from which one obtains 
$\displaystyle \tan\angle BOA={72\over29}$.

*But we also have $\displaystyle {BF'\over OB}={72\over29}$, hence line $OA$ intersects tangent $BF'$ at $F'$, and $F'=F$.

*Now you know $BF={1800\over29}$ and $BG={30\over7}$, thus you can find $FG$ by Pythagoras' theorem.

EDIT.
Notice that $OA$ and $D'E$ meet on the line tangent at $B$ not only in this case, but for any position of point $A$ inside the circle of diameter $BC$. I don't know, however, a simple proof for that.
A: As @Mick suggests in a comment, once you know $\overline{FB}\perp\overline{BC}$, the problem is solved, so I'll just prove that. However, instead of constructing $F$ and showing the perpendicularity property, I'll construct the perpendicular at $B$ and show that it concurs with $\overleftrightarrow{OA}$ and $\overrightarrow{D^\prime E}$, via the trigonometric form of Ceva's Theorem.

Consider $\triangle ABE$ in the figure:

In order to name relevant angles, define $A^\prime$ on $\overleftrightarrow{OA}$, $E^\prime$ on $\overleftrightarrow{D^\prime E}$, and $B^\prime$ on the perpendicular to $\overleftrightarrow{BC}$ at $B$, such that $\overrightarrow{AA^\prime}$, $\overrightarrow{BB^\prime}$, $\overrightarrow{EE^\prime}$ are directed toward the ostensible point of concurrency. (Note: Our diagram and argument assume that $\angle A$ is obtuse; equivalently, that $A$ is inside the circle. The reader is invited to make appropriate adjustments for the acute case, where $A$ is outside the circle.) 
That concurrency is guaranteed if we can show
$$
\frac{\sin\angle A^\prime AE}{\sin\angle A^\prime AB} \;
\frac{\sin\angle B^\prime BA}{\sin\angle B^\prime BE} \; 
\frac{\sin\angle E^\prime EB}{\sin\angle E^\prime EA} = 1 \tag{1}$$ 
Well, consider the following (where I'll write $\angle B$ and $\angle C$ for the angles at those vertices in $\triangle ABC$) ...


*

*$\angle B^\prime BA = 90^\circ - \angle B$, clearly. So, $\sin\angle B^\prime BA = \cos B$.

*$\angle B^\prime B E = \angle C$, as inscribed angles subtending $\stackrel{\frown}{BE}$. So, $\sin\angle B^\prime BE = \sin C$.

*$\angle E^\prime E B = 180^\circ - \angle B$. This follows from the fact that $\angle BED^\prime \cong \angle B$ as inscribed angles subtending congruent arcs $\stackrel{\frown}{CD}$ and $\stackrel{\frown}{BD^\prime}$. So $\sin \angle E^\prime E B = \sin B$.

*$\angle E^\prime EA = 90^\circ +\angle B$. This follows from the additional fact that, by Thales's Theorem, $\angle BEC$ is a right angle. So $\sin \angle E^\prime EA = \cos B$.
Consequently, the left-hand-side of $(1)$ reduces to 
$$
\frac{\sin\angle A^\prime AE}{\sin\angle A^\prime AB} \;
\frac{\cos B}{\sin C} \; 
\frac{\sin B}{\cos B} \qquad\to\qquad
\frac{\sin\angle A^\prime AE}{\sin\angle A^\prime AB} \;
\frac{\sin B}{\sin C}
 \tag{2}$$ 
Now, observe that $\angle A^\prime A B$ is the supplement of $\angle OAB$, whereas $\angle A^\prime AE \cong \angle OAC$. Therefore, $(2)$ becomes
$$\frac{\sin\angle OAC}{\sin C} \;
\frac{\sin B}{\sin\angle OAB}
 \tag{3}$$
where we've arranged the elements in anticipation of invoking the Law of Sines in $\triangle OAB$ and $\triangle OAC$.
$$\frac{|\overline{OC}|}{|\overline{OA}|} \;
\frac{|\overline{OA}|}{|\overline{OB}|} \qquad\to\qquad 1
 \tag{4}$$
Thus, the relation is proven, and Ceva guarantees that the perpendicular at $B$ meets the point of intersection ($F$) of $\overleftrightarrow{OA}$ and $\overleftrightarrow{D^\prime E}$, as claimed. $\square$
