CatalinBordea wrote: Let $ ABC $ be a triangle, $ M_A $ be the midpoint of the side $ BC, $ and $ P_A $ be the orthogonal projection of $ A $ on $ BC. $ Similarly, define $ M_B,M_C,P_B,P_C. M_BM_C $ intersects $ P_BP_C $ at $ S_A, $ and the tangent of the circumcircle of $ ABC $ at $ A $ meets $ BC $ at $ T_A. $ Similarly, define $ S_B,S_C,T_B,T_C. $ Show that the perpendiculars through $ A,B,C, $ to $ S_AT_A,S_BT_B, $ respectively, $ S_CT_C, $ are concurent. Flavian Georgescu It was here. Please use the search function.

Dear Mathlinkers, to begin... 1. ASa is perpendicular to OH Euler's line of ABC 2. Ua the second point of intersection of ASa with the circumcircle of ABC 3. Consider an harmonique pencil with origin Ta 4. who want to continue? Sincerely Jean-Louis

Lemma Let $ABC $ be a triangle ,the altitude and the median $AH,AM$ intersect the ninepoint-circle for second time at respectively $H',M'$ . $ HM,H'M'$ and the tangent to $(ABC)$ at $A$ are concurrent . proof Let $O$ be the circumcenter of $ABC$ , the tangent cuts $BC$ at $T$ then $MTOA$ are cyclic we knows that $AH'MO$is parallelogram and since $AH \parallel OM $ then $AOMK$ is trapezoid where $K=AH\cap (ABC)$.it sclear that $H',K$ are symmetric about $BC$ then $\angle TH'M =\angle MKT =180 ^\circ - \angle TOM$ but $\angle MH'M'=90^\circ -\angle M'MH'$ ,$\angle TOM=90^\circ -\angle MTO$ and $\angle M'MH'=\angle MTO$ i.e $\angle MH'M'=\angle TOM$ hence $\angle TH'M +\angle MH'M'=180 ^\circ$. Back to the problem we have $M_bH_c\cap M_cH_b=S_a$ thus $S_a$ is on the polar of $A$ wrt the $9$ point-circle $(N)$ besides the lemma yields $T$ is on the polar of $A$ wrt $(N)$ hence $NA$ is perpendicular to $TS_a$ idem for the others therefore $AS_a,BS_b,CS_c$ are concurrent at $N$