$BC$ 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

### 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"