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 Geometry problem
Сообщение20.07.2011, 21:53 
Аватара пользователя
It is given a quadrilateral with intersection point of the diagonals - P that is inscribed in circle with center O. M and N are the feets of the perpendiculars from P to AD and BC respectively. K and L are the intersection points of PM with BC and PN with AD respectively. Q is the intersection point of KL and OP. Prove that Q is the middle of KL.

 
 
 
 Re: Geometry problem
Сообщение22.07.2011, 05:04 

(Скрыто модератором zhoraster)

It rubs the lotion on its skin. It does this whenever it's told. It puts the lotion on its skin, or else it gets the hose again.

 !  zhoraster:
Замечание за оффтоп!

 
 
 
 Re: Geometry problem
Сообщение22.07.2011, 07:36 
Оказывается, центры описанных окружностей треугольников $APD$ и $BCP$ лежат на прямых $NP$ и $MP$ соответственно. Поэтому $OEPF$ - параллелограмм и остается показать, что прямые $FE$ и $KL$ параллельны.

Откуда задача?

 
 
 
 Re: Geometry problem
Сообщение23.07.2011, 00:19 
Аватара пользователя
Thank you for the valuable idea and the time spent. A friend of mine said that is not a new problem but not said the exact source where he saw the problem. I discovered it on my own as a fact using software. I like discovering new statements and finding a solution for them. I hope you like the problem.

(Оффтоп)

/I'm sorry for my bad English but my Russian is worst./

 
 
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