# 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 :(

• * Application Commented Sep 21, 2017 at 13:10
• Under what circumstances you can apply the power of a point? Is it because FB is tangent to BDCD'?
– Mick
Commented Sep 21, 2017 at 14:49
• oh I mean I am applying power of a point at the beginning to get $AE/AD=17/12$ and I tried to use Ptolemy's theorem on both BDCD' and BDCE but they don't work Commented Sep 21, 2017 at 14:57
• Have you ever thought of $\angle FBO = 90^0$?. If that is true, the whole problem can be solved.
– Mick
Commented Sep 21, 2017 at 14:59
• wait how do you get there? is it by angle chasing? Commented Sep 21, 2017 at 15:09

Here's a scheme of the solution.

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

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

3. 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: $\displaystyle BF'={1800\over29}$.

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

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

6. 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.

• wait i think cos(abc) should be 4/5 though Commented Sep 21, 2017 at 21:11
• Right, thank you: corrected now. Commented Sep 21, 2017 at 21:12
• and do you mean sinACB= 72/170 instead of sinABC=72/170? Commented Sep 21, 2017 at 21:15
• Yes, of course: just another typo. Commented Sep 21, 2017 at 21:17
• and can you just show how do you arrive at CD=BD'=30? I tried law of sines but it doesm't work. Commented Sep 21, 2017 at 21:18

It looks that Stewart’s Theorem is useful here.

It states that in $\triangle ABC$ \begin{align} a^2\,n+b^2\,m&=c\,(d^2+m\,n) . \end{align}

\begin{align} |CD|&=|BC|\,\sin\beta =30 ,\\ |BG|&=\tfrac17|CD|=\tfrac{30}7 ,\\ |OG|&=\tfrac12|BC|-|BG|=\tfrac{145}{7} ,\\ |BE|&=|BC|\,\sin\gamma =\tfrac{360}{17} ,\\ |BD|&=|BC|\cos\beta=40 ,\\ |CE|&=|BC|\cos\gamma=\tfrac{770}{17} ,\\ |AE|&=|CE|-|AC|=\tfrac{192}{17} ,\\ \angle ED'D &=\tfrac\pi2-\beta-\gamma ,\\ |DE|&=|BC|\,\sin\angle ED'D \\ &=|BC|\,\sin(\tfrac\pi2-\beta-\gamma) \\ &=|BC|\,\cos(\beta+\gamma) \\ &=|BC|\,(\cos\beta\cos\gamma-\sin\beta\sin\gamma) \\ &=|BC|\,\left(\sqrt{(1-\sin^2\beta)(1-\sin^2\gamma)}-\sin\beta\sin\gamma\right) \\ &=|BC|\,\tfrac8{17} =\tfrac{400}{17} ,\\ |D'E|&=\sqrt{|BC|^2-|DE^2|} =\tfrac{750}{17} ,\\ \end{align}

According to Stewart’s Theorem, in $\triangle ABC$

\begin{align} |OA|&= \sqrt{\tfrac12(|AB|^2+|AC|^2)-\tfrac14|BC|^2} =\sqrt{241} ,\\ |AM|&=25-\sqrt{241} . \end{align}

According to Stewart’s Theorem, in $\triangle OME$

\begin{align} |EM|&= \sqrt{\frac{|OM|(|AE|^2+|OA||AM|)-|OE|^2|AM|}{|OA|}} =\tfrac5{4097}\sqrt{839270450-51431810\sqrt{241}} . \end{align}

According to Stewart’s Theorem, in $\triangle ODM$

\begin{align} |MD|&= \sqrt{\frac{|OM|\,(|AD|^2+|OA|\,|AM|)-|OD|^2|AM|}{|OA|}} =\tfrac5{241}\sqrt{2904050-147010\sqrt{241}} . \end{align}

According to Stewart’s Theorem, in $\triangle BOM$

\begin{align} |BM|&= \sqrt{\frac{\tfrac12|BC|(|AB|^2+|OA|\,|AM|)-1/4*BC^2*AM}{|OA|}} =\tfrac5{241}\sqrt{2904050-69890\sqrt{24}} . \end{align}

Also, \begin{align} \cos\angle BOM&= 1-2\sin^2\tfrac12\angle BOM =1-2\left(\frac{|BM|}{|BC|}\right)^2 =\tfrac{29}{1205}\sqrt{241} . \end{align}

According to the power of a point $F$,

\begin{align} \frac{|EF|}{|NF|} &= \frac{|MF|}{|D'F|} =k \tag{1}\label{1} , \end{align}

hence $\triangle MFE \sim \triangle D'FN$ and \begin{align} k& =\frac{|EM|}{|D'N|} =\frac{|EM|}{|MD|} =\tfrac5{102}\sqrt{241}-\tfrac{29}{102} . \end{align}

Now \eqref{1} can be rewritten as a system of two equations in two unknowns, $|EF|$ and $MF$:

\begin{align} \frac{|EF|}{|MF|+|BC|} &=k ,\\ \frac{|MF|}{|EF|+|D'E|}&=k , \end{align}

it follows that \begin{align} |MF|&= \frac{k\,(k\,|BC|+|D'E|)}{1-k^2} =\tfrac{125}{29}\sqrt{241}-25 ,\\ |EF|&= \frac{k\,(k\,|D'E|+|BC|)}{1-k^2} =\tfrac{21600}{493} ,\\ |OF|&=|MF|+\tfrac12|BC|=\tfrac{125}{29}\sqrt{241} . \end{align}

Finally, from $\triangle FGO$, \begin{align} |FG|&= \sqrt{ |OG|^2+|OF|^2-2|OG|\,|OF|\cos\angle BOA } =\frac{12630}{203} . \end{align}

As a bonus, \begin{align} |BF|^2&= \tfrac14|BC|^2+|OF|^2-|BC|\,|OF|\cos\angle BOA =\tfrac{3240000}{841} ,\\ |FG|^2-|BG|^2 &=\tfrac{3240000}{841} , \end{align}

so, indeed, $BF\perp BC$.

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. $\square$

• wait will Pascal's theorem give us the concurrency immediately? Commented Sep 23, 2017 at 12:05
• @Guywhofailedeverything: I'm not seeing it.
– Blue
Commented Sep 23, 2017 at 17:22