$BC$ is a diameter of a circle and the points $X,Y$ are on the circle such that $XY \perp BC$.The points $P,M$ are on $XY,CY$(or their stretches),respectively,such that $CY \parallel PB$ and $CX \parallel PM$.Let $K$ be the meet point of the lines $XC,BP$.Prove that $PB \perp MK$.
(An easy problem.Just a little angle chasing)
Iran NMO 2005/2
"Questions we can't answer are far better than answers we can't question"
Re: Iran NMO 2005/2
are you sure about the statement?
this line is confusing.tanmoy wrote: $CX \parallel PM$.
Re: Iran NMO 2005/2
opsssss sorry my mistake
here is the solution:
$PM=YM$ [because $PM||XC$]
so,$KC=YM$
$\angle KBC=\angle BCY=\angle KCB$
$\therefore KB=KC$
$KB=KC=MY$
$\therefore MYBK$ is a parallelgram.
$\therefore \angle MKB=\angle MYB =90$
so, $MK$ is perpendicular to $PB$ .
here is the solution:
$PM=YM$ [because $PM||XC$]
so,$KC=YM$
$\angle KBC=\angle BCY=\angle KCB$
$\therefore KB=KC$
$KB=KC=MY$
$\therefore MYBK$ is a parallelgram.
$\therefore \angle MKB=\angle MYB =90$
so, $MK$ is perpendicular to $PB$ .
Re: Iran NMO 2005/2
Same as my solution.Tahmid wrote:opsssss sorry my mistake
here is the solution:
$PM=YM$ [because $PM||XC$]
so,$KC=YM$
$\angle KBC=\angle BCY=\angle KCB$
$\therefore KB=KC$
$KB=KC=MY$
$\therefore MYBK$ is a parallelgram.
$\therefore \angle MKB=\angle MYB =90$
so, $MK$ is perpendicular to $PB$ .
"Questions we can't answer are far better than answers we can't question"