A Helpful Generalization (Self-Made)

[Phlembac Adib Hasan]

I've made generalization of my lemma given in viewtopic.php?f=25&t=1694.

Let $\triangle ABC$ and $\triangle ADE$ are spiral symmetric.Let $M$ is any point of the plane.Let $N$ is the corresponding point of $M$ comparing $M$ to $\triangle ABC$ and $N$ to $\triangle ADE$.(That is $M$ and $N$ are any two points on the plane such that $\frac{AM}{AN}=\frac{AB}{AD}=\frac{AC}{AE}=\frac{BC}{DE}$).Prove that $\angle ABD=\angle ACE=\angle AMN $ and $\angle ADB=\angle AEC=\angle ANM$.I'll say it "Spiral Homothety" .Go to the link given above to see its application.
Hint :

Notice that $\triangle ABD, \triangle ACE, \triangle AMN\; \; $ are spiral symmetric.

[*Mahi*]

Proof :
Let the center of rotation be $0$ and the ratio by which it is rotated be $\alpha$. Then every point $x$ is mapped to $\alpha x$ under spiral similarity. Then the angle between center of rotation, a point and corresponding point after rotation will be a constant, varying for every combination those are taken.
(Those can be argument of \[\frac{1}{1-\alpha},\frac{1}{\alpha},\frac{\alpha}{1-\alpha}\]) or their multiplicative/additive inverse.)

[*Mahi*]

The lemma statement can be stated like this:

Center of Spiral Symmetry $O$, a point $A$, and it's respective point $A'$ always form a triangle with equal angle measures, which depends only on the ratio and angle of rotation .

[Phlembac Adib Hasan]

হেব্বি গাণিতিক গাণিতিক লাগতেসে । I like.