Nice but not easy !

[stergiu]

Let ABC an acute triangle. The circle with diameter the altitude BD meets sides

AB,BC at K,L respectively. The tangents of this circle at K,L meet at M.Prove that

line BM bisects segment AC.

Babis

[karlosson_sul_tetto]

Well, I did a drawing and I think watching it is always possible, with ABC acute:

Beh, ho fatto un disegno e penso guardandolo che è sempre possibile, con la ABC acutangolo:

Ну,я сделал рисунок и смотря на него думаю что всегда возможно:



Bueno, hice un dibujo y creo que viendo que siempre es posible, con ABC aguda



E pra, unë bëra një vizatim dhe unë mendoj se shikuar se është gjithmonë e mundur, me ABC akut



Nou, ik heb een tekening te kijken en ik denk dat het altijd mogelijk, met ABC acute

Nå, jeg gjorde en tegning, og jeg tror, at se det er altid muligt, med ABC akut



Λοιπόν, έκανα ένα σχέδιο και νομίζω ότι βλέποντας ότι είναι πάντα δυνατό, με οξεία ABC

Nun, ich habe eine Zeichnung und ich denke, gerade ist es immer möglich, mit ABC akuten



Eh bien, j'ai fait un dessin et je pense à regarder, il est toujours possible, avec ABC aiguë

[Tibor Gallai]

Non ho capito cosa è sempre possibile...
Comunque il problema chiede di dimostrare un'affermazione.

[karlosson_sul_tetto]

Be,io ho detto che ad occhio è sempre possibile.

[¬[ƒ(Gabriel)³²¹º]¼+½=¾]

@karlosson: lol

btw let H be the projection of B on CA, so $ \angle LKB = \angle LHB = 90 - \angle LBH = \angle BCA $ and equally we have $ \angle KLB = \angle CAB $, so BLK and BCA are similar. Olso the simmetric of BLK wrt the angle bisector of $ \angle ABC $ is a triangle homotetic with ABC, with an homothety with center B. But BM is the symmedian of BKL as know, so it's also the median of ABC.

I think that this prof is valid also if the triangle is not acute.