By simply angles we can prove that triangles MDA, MBC, SDC and TBA are similar. Thus we have: MS=BD*CM/BC=(AC*BM/CM)*(CM/BC)=AC*BM/BC=MT PS: What is CGMO?

CGMO = Chinese Girls Mathmetical Olympiad. It is a contest only for girls. The problems are relatively easy.

My solution: Since $ M $ is the Miquel point of complete quadrilateral $ \{ AC, AD, BC, BD \} $ , so we get $ \triangle MCA \sim \triangle MBD $ and $ \frac{BM}{CM}=\frac{BD}{AC} $ . ... $ (1) $ Since $ \triangle BTM \sim \triangle BAC $ , so we get $ \frac{TM}{AC}=\frac{BM}{BC} $ . ... $ (2) $ Since $ \triangle CMS \sim \triangle CBD $ , so we get $ \frac{MS}{BD}=\frac{CM}{BC} $ . ... $ (3) $ From $ (1), (2), (3) $ we get $ TM=MS $ . Q.E.D

Dear Mathlinkers, you can see and more http://jl.ayme.pagesperso-orange.fr/ vol. 16 Deux triangles adjacents..., p. 57-61. Sincerely Jean-Louis

Apply an inversion around $P$: then $\{P\}= A'C' \cap B'D'$, $\{M'\}= A'D' \cap B'C'$, $\{T'\}=A'B' \cap PM'$ and $\{S'\}=C'D' \cap PM'$. Since $(M',P;T',S')=-1$, it follows that $$\frac{MT}{MS}=\frac{\frac{M'T'}{T'P}}{\frac{M'S'}{S'P}}=1.$$

Inverse around $O$,then $M'=A'D'\cap B'C',T'=OM'\cap A'B',S'=OM'\cap C'D'$. Since inversion preserves cross ratio,we have $$(T,S;M,\infty_{TS})=(T',S';M',O)=-1$$Done.