We complex bash: (Will insert Part a) later )

Part a) is Russia 9.2. 2013

a) By $SSS$, $\triangle TMB \cong \triangle TMC$, so $\angle TMB = \angle TMC = 90^\circ$. Since $\angle TEC = 90^\circ, TMCE$ is cyclic. Let $EM$ meet $AB$ at $F$. Then, $\angle FEA = \angle FEC = \angle MEC = \angle MTC = 90^\circ - \angle TCM = 90^\circ - \angle TCB = 90^\circ - \angle BAC = 90^\circ - \angle FAE$. So, $\angle AFE = 90^\circ$. Similarly, if we let $DM$ meet $AC$ at $G$, we can get that $\angle AGD = 90^\circ$. Thus, $EF$ and $DG$ are altitudes and their intersection, $M$, is the orthocenter of $\triangle ADE . \square$ b) From a), $\angle ADG = \angle AEF = 90^\circ - \angle BAC$. So, $\angle MDT = \angle MET = \angle BAC$. Also, since $AFMG$ and $ADTE$ are cyclic, $\angle DME = \angle FMG = 180^\circ - \angle BAC$ and $\angle DTE = 180^\circ - \angle BAC$. Thus, $DMET$ is a parallelogram and $TM$ bisects $DE. \square$

MathsLion wrote: Part a) is Russia 9.2. 2013 and hence it also is EGMO 2023 P2

We can easily see that TDBM and TECM are cyclic. Then ∠TEM=∠TCM=∠TBC=∠TDM (Then ∠TEM=∠TDM (1) ). And ∠TMD+∠TDM=∠TBD+∠TBM=∠C+∠BAC=∠MTE+∠BAC (2), then from (2) and ∠TBM=∠BAC we obtain that ∠TMD=∠TBD=MTE (3). From (1) and (3) we have TDM and MET triangles are congruent (they are similar from (1) and (3), and congruent from MT is common in these triangles0. Then MD=TE and DT=ME which means that DTEM is parallelogram (4). From (4) we have ME // DT and MD//TE which means that EM is perp. to AD and DM is perp. to AE, which means that M is orthocenter of ADE and from (4) we know that MT cuts DE at midpoint of DE, which completes the part a and b.

For part b) just prove that; $TM \cap DE=K$ therefore $\frac{tm(d+e)-de(t+m)}{tm-de}=^{?}\frac{d+e}{2}$ where $2m=b+c$ $t(b+c)=2bc$ $2d=a+b+t-ab/t$ $2e=a+c+t-ac/t$ Where $(DEF)\in \mathbb{S^1}$