In the following figure prove that $AE||EF$ (without using similar triangles)...

Question

From an arbitrary point $A$ outside a circle draw a tangent $AB$ to the circle and select point $E$ also outside the circle such that $AB=AE$.Let $D$ be an arbitrary point on perimeter of circle , call the other intersection of $AD$ with circle as $C$ . The other intersection points of $ED$ , $EC$ with circle are called $G$ , $F$ respectively.Without using triangle similarity prove that: $AE||FG$

I know there is a fairly simple solution using similar triangles,But is there any other method to solve this??

Recently after hours of works I made some progress for solving this question as seen in the following figure:

Now its sufficient to prove that $MG=MF$ , but can't go further!

Answer 1

The key step is to realize that the unique circle through C, D, E, is tangential to AE. Based on that fact, it is a simple angle chasing to show that

$$ \angle AED = \angle ECD = \angle FCD = \angle FGD $$

and hence the 2 lines are parallel.

How do we show the fact? Well, by power of a point on the given circle, we know that $AD \times AC = AB^2 $. Since $AB = AE $, hence $AD \times AC = AE^2 $. Thus, the circle through C, D, and E is indeed tangential to AE.

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Note: Not sure what defines "Without using triangle similarity". In particular, power of a point is essentially triangle similarity.

Answer 2

Does inversion count?

Let us name the given circle $c_1$ and let $O$ be its center. Draw circle $c_A$ centered at $A$ and with radius $AB = AE$. Then $c_A$ passes through points $B$ and $E$. Since $AB$ is tangent to $c_1$, we have $\angle \, OBA = 90^{\circ}$ which means that circle $c_1$ is orthogonal to circle $c_A$. Then if we perform inversion with respect to $c_A$, circle $c_1$ is mapped to itself because it's orthogonal to $c_A$. Therefore point $C$ is mapped to point $D$ under this inversion.

Now, draw the unique circle $c_2$ through points $C, D, E$. Under the inversion with respect to $c_A$, point $C$ is mapped to point $D$ and therefore the circle $c_2$ is mapped by the inversion to itself. Hence circle $c_2$ is orthogonal to circle $c_A$. This means that if $O_2$ is the center of $c_2$ then $\angle \, O_2EA = 90^{\circ}$ which yields that $AE$ is tangent to $c_2$ at point $E$.

Finally, $\angle \, FGD = \angle \, FCD$ because points $C, F, D, G$ lie on the common circle $c_1$. Thus $$\angle \, FGE = \angle \, FGD = \angle \, FCD = \angle \, ECD .$$ As we already saw, $AE$ is tangent to circle $c_2$ at point $E,$ so $\angle \, ECD = \angle \, AED = \angle \, AEG$. Consequently, $\angle \, FGE = \angle \, AEG$ which is equivalent to the fact that $FG$ is parallel to $AE$.