Prove $AF, BC, GE$ are concurrent in a tangent–circumcircle configuration

Question

Problem: enter image description here

Let $\triangle ABC$ be inscribed in a circles. $D$ be the intersection of the tangents at $B$ and $C$. Let $AD$ intersect the circumcircle at $E$. Let $F$ be any point on the circumcircle on the arc which contains $A$ and $AF$ is not parallel to $BC$. Let $DF$ intersect the circumcircle at $G$. Prove that $AF$,$BC$ and $GE$ are concurrent.

My approach: I defined $K'$ as the intersection of $Af$ and $BC$, and tried to show $E$,$G$ and $K'$ are collinear using angle chasing. But I couldn't extract much information using angle chasing. So, I switched to length chasing but eventually failed. Also, I have an observation. In $\triangle ADF$, $E$,$F$ and $K'$ lie on the sides of the triangle. Since we are trying to prove they are collinear, we can have a target expression using Menelaus theorem. I tried to chase that backward but couldn't do anything. Last observation, $AD$ is the symmedian of $\triangle ABC$.

Background: I was working on a different contest problem but got this result as a subproblem. Then, I tried to only prove this but failed.

Conclusion: As an answer, I would prefer some hints and solution in hidden format. So that I can check hints(but not see the rest of the hints) or peep at a chuck of solution if needed. Also, what kind of bash would be helpful here? In contest, if I am stuck, I have to somehow solve. So, bash can work as a backup plan. I tried coordinate and it went horribly wrong. At last, I would be glad if you share your intuition. Any suggestion is appreciated.

EDIT: I want a purely Euclidean(no pole and polar,inversion,homothety, projective geometry), or purely bashed(complex numbers,coordinate,baycentric coordinate) solution.

• RobinSparrow: I suspect you can use Pascals theorem here.
• Math12: Isn't pascal theorem used to prove 3ドル$ points colinear inside the circle? Here, the concurrency point is outside the circle. If you have a way to use pascal's theorem or any idea, feel free to share.
• Geometry99: How about desargues theorem on $\triangle ABE$ and $\triangle FCG$?
• Math12: Anyone?????????

Answer 1

Define $K$ as $AF\cap GE$. We wish to show $K$ lies on line $BC$.

By Brocard's theorem on $AFGE$, $K$ lies on the polar of $D$, which is line $BC$. We are done.

• Math12: What is Brocard's theorem? I didn't read any theorem like it in A beautiful Journey Through Olympiad Geometry book.
• CosmicOscillator: @Math12 See en.wikipedia.org/wiki/Brokard%27s_theorem. It is also covered in the book Euclidean Geometry in Math Olympiads in the section on projective geometry. I recommend learning projective geometry since it is very useful in dealing with certain configurations.
• Math12: Isn't there any way to solve it using Euclidean geometry since I don't know inversion, pole-polar and projective geometry. Won't knowing $AD$ is the symmedian work?
• Math12: Isn't there any way using synthetic geometry?
• Gorg: @Math12, if you really want to advance in elementary geometry, you'll have to study all these topics at some point. These techniques are fundamental in modern elementary geometry, and your original question is essentially nothing but a basic fact about poles and polars.

Answer 2

Let $X = AF \cap BC$ and $Y = GE \cap BC$. We will prove that $X$ and $Y$ coincide by showing that they divide $BC$ in the same ratio: $$\frac{BX}{XC} = \frac{BY}{YC}$$

Applying Sine Rule to triangles $\triangle ABX$ and $\triangle ACX$: $$BX = AB \cdot \frac{\sin(\angle BAX)}{\sin(\angle AXB)}, \quad XC = AC \cdot \frac{\sin(\angle CAX)}{\sin(\angle AXC)}$$ Since $B, X, C$ are collinear, $\angle AXB$ and $\angle AXC$ are supplementary, so their sines cancel. Using the Extended Sine Rule, replacing sines with chords: $\sin(\angle BAX) = \frac{BF}{2R}$ and $\sin(\angle CAX) = \frac{CF}{2R}$. $$\frac{BX}{XC} = \frac{AB}{AC} \cdot \frac{BF}{CF} \tag{1}$$

Now we apply the same logic to point $Y$.

Applying the Sine Rule to $Y = GE \cap BC$: $$\frac{BY}{YC} = \frac{GB}{GC} \cdot \frac{\sin(\angle BGY)}{\sin(\angle CGY)}$$ The angles subtend chords $BE$ and $CE$. $$\frac{BY}{YC} = \frac{GB}{GC} \cdot \frac{BE}{CE} \tag{2}$$

We need to relate the chords $BE, CE$ back to $AB, AC$ to compare the two equations.

Since $DB$ and $DC$ are tangents, $\triangle DBE \sim \triangle DAB$ and $\triangle DCE \sim \triangle DAC$. This implies $\frac{BE}{AB} = \frac{DB}{DA}$ and $\frac{CE}{AC} = \frac{DC}{DA}$. Dividing these (and using $DB=DC$): $$\frac{BE}{CE} = \frac{AB}{AC} \tag{3}$$ Similarly for the other secant $DGF$: $$\frac{GB}{GC} = \frac{FB}{FC} \tag{4}$$

Substitute (3) and (4) into equation (2) to get: $$\frac{BY}{YC} = \frac{FB}{FC} \cdot \frac{AB}{AC}$$ This matches equation (1). Since $\frac{BX}{XC} = \frac{BY}{YC}$, points $X$ and $Y$ are identical. Thus, $AF, GE, BC$ are concurrent.

• Math12: How did you get the idea of trigonometry and taking ratios? I was just chasing angles and maybe some ratios related to Ptolemy, intersecting chords theorem. I want to know your intuition.
• 匚ㄖㄥᗪ乇ᗪ: @Math12 This may be a disappointing answer but I am not really good at geometry and I haven't had any formal training, so the stuff in the comments like poles and all are not familiar to me so I just tried with what elementary knowledge I have, and supposedly it worked.

Answer 3

Since the OP wants a set of hints involving no pole-polars, I'll just reverse engineer it. I will use Pascal's Theorem as the OP has shown familiarity in the comments.

Lemma 1:

Given a triangle $ABC$ and its symmedian $AE$, tangent to its circumcircles at $A$ and $E$ meet on $BC$.

Proof:

You probably know that $ABED$ is harmonic, so you're done. If not, drudge along: Show that the condition is equivalent (and not just implies) $AB/AC = EB/EC$ by simple applications of sine rule. Note this rearranges to give $BA/BE = CA/CE$, which, using the same equivalence, gives the lemma.

Define $K' = AG \cap EF$. Now, we can finish as follows:

Define the intersection of $A$-tangent and $E$-tangent as $X$, and intersection of $F$-tangent and $G$-tangent as $Y$. Use Pascal's theorem on $AAFEEG$, which gives $X,K,K'$ are collinear. Use Pascal's theorem on $FFAGGE$, whcih gives $Y,K,K'$ are collinear. Combining, $K,K',X,Y$ are collinear. But line $XY$ is the same as line $BC$ (by application of Lemma to $ABC$ and $FBC$), so we're done.