Let $ABC$ be a triangle such that $|AB|=|AC|$ and let $D,E$ be points on the minor arcs $\overarc{AB}$ and $\overarc{AC}$ respectively. The lines $AD$ and $BC$ intersect at $F$ and the line $AE$ intersects the circumcircle of $\triangle FDE$ a second time at $G$. Prove that the line $AC$ is tangent to the circumcircle of $\triangle ECG$.
Problem
Source: Turkey TST 2015
Tags: TST, geometry, geometry proposed, Inversion, tangent
01.04.2015 11:38
My solution: Since $ BC $ is the image of $ \odot (ABC) $ under Inversion $ \mathbf{I}(A, AB) = \mathbf{I}(A, AC)=\Psi $ , so $ D \longleftrightarrow F, E \longleftrightarrow AE \cap BC $ under $ \Psi \Longrightarrow G \in BC \Longrightarrow AG \cdot AE=AC^2 \Longrightarrow AC $ is tangent to $ \odot (CEG) $ at $ C $ . Q.E.D
01.04.2015 12:48
My sol is similar to TelvCohl, but my last step: Let $AC$ intersect $\odot (CEG)$ at $C'$. Then, as the inverse of $\odot (CEG)$ under $\Psi$ is $\odot (CEG)$ itself and as $C$ lies on $\odot (CEG)$, it follows that the inverse of $C$ under $\Psi$ is $C'$ or the inverse of $C'$ under $\Psi$ is $C'$ itself. In either case, $C'$ coincides with $C$.
01.04.2015 13:27
Any synthetic solution, not including inversion?
01.04.2015 13:37
G is actually on BC. You can prove it with angle chasing. Then some more to finish the problem.
01.04.2015 15:25
navredras wrote: Any synthetic solution, not including inversion? Solution: $ ABC=ACB=BDF$ and $ABC+DBA=BDF+AFC=ABC+AFC $ so $ AFB=ABD $ and we get that $ AB $ is tangent to circumcircle of $\triangle BDF $ so $ AB^2=AD \cdot AF=AC^2=AE \cdot AG $ DONE
01.04.2015 16:14
Well, this is pretty much the same solution as TelvCohl's and andria's ones. By inversion, $AD \cdot AF = AB^2$. From the power of A : $ AD \cdot AF=AE \cdot AG$. Since $AB=AC$, we obtain $AC^2=AE \cdot AG$, so $AC$ is tangent to $\odot CEG$.
04.04.2015 13:08
To see $G$ lies on $BC$, let $ AE \cap BC = \{K\} $. $\angle ADE = \angle ACE = \alpha$, $\angle CAE = \beta \Longrightarrow \angle ABC = \angle ACB = \alpha + \beta$. Hence $\angle AKF = \alpha + \beta - \beta = \alpha$. $\angle AKF = \angle ADE = \alpha$, which means $FDEK$ is cyclic so $K$ coincides with $G$. Since $\triangle ACE \sim \triangle AGC$ from $AA$ similarity, we get $|AC|^2 = |AE| \cdot |AG|$. This finishes the proof.
30.04.2015 10:47
I have a proof by contradiction.......We are supposed to prove that $|AC|^2 = |AE|\cdot|AG|$ or that is we have to prove $|AB|^2 = |AD|\cdot|AF|$ which is equivalent to prove that $\angle DBA = \angle BFA = \angle DEA$.....or it trivially follows to prove that $G$ lies on the line $BC$.....assume the contrary then apply Pascal to the degenerate hexagon namely $FDEGCB$ to get $ FC \cap BD $, $ DG \cap EC $ and $ FG \cap EB $ to be collinear and thus we arrive at a contradiction.... Q.E.D
21.02.2017 06:52
$\angle ABC = \angle ACB = \angle FDB$ so $AE \cdot AG = AD \cdot AF = AB^2 = AC^2$
05.01.2018 15:38
Nice solution !
02.12.2019 21:51
Just Posting for storage.... Turkey TST 2015 wrote: Let $ABC$ be a triangle such that $|AB|=|AC|$ and let $D,E$ be points on the minor arcs $\overarc{AB}$ and $\overarc{AC}$ respectively. The lines $AD$ and $BC$ intersect at $F$ and the line $AE$ intersects the circumcircle of $\triangle FDE$ a second time at $G$. Prove that the line $AC$ is tangent to the circumcircle of $\triangle ECG$. Solution:- Construct a circle $\omega$ with raduis $AB$ centered at $A$, obviously $C\in\omega$. So, $\Psi_{\omega}:(\odot(ABC))\mapsto BC$. Hence, $\Psi_{\omega}:D\mapsto F$. So, $AD.AF=AE.AG=AB^2=AC^2\implies AC$ is tangent to $\odot(ECG)$.
02.12.2019 23:39
aloski1687 wrote: Let $ABC$ be a triangle such that $|AB|=|AC|$ and let $D,E$ be points on the minor arcs $\overarc{AB}$ and $\overarc{AC}$ respectively. The lines $AD$ and $BC$ intersect at $F$ and the line $AE$ intersects the circumcircle of $\triangle FDE$ a second time at $G$. Prove that the line $AC$ is tangent to the circumcircle of $\triangle ECG$. Clearly by angle Chasing $AB$ is tangent to $\odot{BDF}$.Hence \[AE\cdot AG=AD\cdot AF=AB^2=AC^2 \implies \odot{ECG} \text{tangent to} AC\].
03.12.2019 04:07
This image is highly symmetric One can argue that F, B and G are collinear and easy angle chase shows that AB is tangent to DBF, then since the diagram is symmetric, AC is tangent to ECG
31.03.2020 18:57
So first of all let me define an inversion $\psi(A,\sqrt{AB.AC})=\psi(A,AB)$ By this inversion point $D$ maps to point $F$, by the power of point we have that $AD.AF=AE.AG$. But by inversion we know that $AD.AF=AB^2=AC^2$, thus we have that $AE.AG=AC^2$. But if we watch closely we now know form this that the points $F,B,C,G$ are all collinear. Because from $AE.AG$ we got that from $E$ maps to $G$ under the inversion $\psi$ and also now we can consider the power of point $A$ with respect to the circle around $\triangle ECG$, we have that $AE.AG = AC^2$. Now this means that $AC$ is tangent to the circle......
14.11.2021 11:03
By performing an inversion with centre $A$ and radius $AB$ we know that $AD \cdot AF = AB^2$ Now let $BC \cap \odot(FDE) = G'$ $$\measuredangle DEG' = \measuredangle DFG' = \measuredangle DFB = \measuredangle DBA = \measuredangle DEA \implies A-E-G' \implies G' \equiv G$$Now $$AD \cdot AF = AB^2 = AC^2 = AE \cdot AG \ \blacksquare$$
14.11.2021 18:39
Let $\angle ABC=\angle ACB=\angle FDB$ Let $\angle DFB=\angle ABD$ $\implies \triangle ADB \simeq \triangle ABF$ $\implies AD.AF=AB^2$ but $AB^2=AC^2$ by $FDEQ$: $AD.AF=AE.AG$ Then: $AD.AF=AB^2=BC^2=AE.AG$ this means that $AC$ is tangent to the circle $\odot{ECG}$
23.12.2021 07:35
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