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The construction of figures to produce special results in plane geometry is an interesting problem, attracting the attention not only of students but also of professional mathematicians. Pizza's theorem, Napoleon's theorem, Butterfly theorem, Japanese theorem, Morley's trisector theorem, Thebault's theorem... are some typical examples of theorem statements. I found a nice configuration as follows. I think this result can be considered the best generalization of Bottema's theorem to date. I am looking for poof of this result:

Given triangle $ABC$, $A'$ is any point in the plane, $M$ is an arbitrary point on $BC$. Construct triangles $\triangle ACB'$ similar to triangle $\triangle BMA'$, triangle $\triangle ABC'$ similar to triangle $\triangle CMA'$, then $A', B', C'$ are collinear and

$$\frac{A'C'}{A'B'}=\frac{AC.BC'}{AB.CB'}=\frac{MB}{MC}$$

enter image description here

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  • $\begingroup$ A simple computational proof is as follows: we use complex coordinates $0,1,z,w,x$ for $B,C,A,A',M$ ($x$ being real). We can easily compute the coordnates of $B',C'$ from the data and thus verify the result $\endgroup$ Commented Jul 10 at 5:56
  • $\begingroup$ This method also allows a quantitative version, i.e., it gives a formula for the area of the triangle $A'B'C'$ in the general case ($M$ not necessarily on the line $BC$), the expression vanishing precisely when that is indeed the case (thus giving a kind of converse). $\endgroup$ Commented Jul 10 at 7:00
  • $\begingroup$ Can I see your figure? $\endgroup$ Commented Jul 10 at 7:16
  • $\begingroup$ Same as yours except that $M$ can be anywhere in the plane. $\endgroup$ Commented Jul 10 at 8:27
  • $\begingroup$ If M be any point in the plane, are A', B', C' still collinear? @terceira $\endgroup$ Commented Jul 10 at 8:46

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It is a typical problem about composition of rotational homotheties (RH).

Let $H_b$ be a RH with center $B$ which maps $C'$ to $A$, $H_c$ be a RH with center $C$ which maps $A$ to $B'$. Then the composition $H:=H_c(H_b)$ is RH which maps $C'$ to $B'$, has coefficient $\lambda:=(BA/BC')\cdot (CB'/CA)$ and angle $\angle C'BA+\angle ACB'=\pi$. Denote by $A''$ the center of $H$, then $A''$ lies on the segment $C'B'$ and enjoys the relation $A''B'/A''C'=\lambda$. Denote $K=H_b(A'')$, then $H_c(K)=H(A'')=A''$. It yields that $\angle A''BK=\angle C'BA, \angle A''CK=\angle B'CA$, thus $BA''CK$ is a cyclic quadrilateral. Next, $\angle A''CB=\angle A''KB=\angle C'AB=\angle A'CB$, i.e., $A'',A',C$ are collinear. Analogously, $A'',A',B$ are collinear, and we have $A'=A''$. It remains to note that from similarities $MB/MC=(MB/MA')\cdot (MA'/MC)=(CA/CB')\cdot(BC'/BA)$.

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  • $\begingroup$ Can you proof $\frac{A'C'}{A'B'}=\frac{MB}{MC}$ ? $\endgroup$ Commented Jul 10 at 6:27
  • $\begingroup$ This is proved, both are equal to $1/\lambda$ $\endgroup$ Commented Jul 10 at 6:43
  • $\begingroup$ What is a rotational homothety? $\endgroup$ Commented Jul 10 at 11:19
  • $\begingroup$ @IosifPinelis a composition of roration and homothety with the same center. Equivalently, a map $z\to az+b$ on the complex plane with $a\ne 0$ (and $a\ne 1$, but it is natural not to exclude parallel translations). $\endgroup$ Commented Jul 10 at 16:44
  • $\begingroup$ @FedorPetrov : Thank you for your response. $\endgroup$ Commented Jul 10 at 17:08
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For the collinearity, this is equivalent to Pascal theorem. See figure: red angles are supplementary and the other same colored angles are equal.

Pascal colin

This is for collinearity as the given conditions give six points on the same circle $(ABC)$. We then apply Pascal theorem to ordered hexagon $AECDBF$. For the ratios (last one), this needs some coordinate computation (or reasoning) in the plane as in the answers given.

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