Difference between revisions of "2006 Romanian NMO Problems/Grade 9/Problem 2"

Revision as of 10:14, 28 July 2006

Problem

Let $\displaystyle ABC$ and $\displaystyle DBC$ be isosceles triangle with the base $\displaystyle BC$ . We know that $\displaystyle \angle ABD = \frac{\pi}{2}$ . Let $\displaystyle M$ be the midpoint of $\displaystyle BC$ . The points $\displaystyle E,F,P$ are chosen such that $\displaystyle E \in (AB)$ , $\displaystyle P \in (MC)$ , $\displaystyle C \in (AF)$ , and $\displaystyle \angle BDE = \angle ADP = \angle CDF$ . Prove that $\displaystyle P$ is the midpoint of $\displaystyle EF$ and $\displaystyle DP \perp EF$ .

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 ==See also==
 *[[2006 Romanian NMO Problems]]
+[[Category:Olympiad Geometry Problems]]

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Difference between revisions of "2006 Romanian NMO Problems/Grade 9/Problem 2"

Revision as of 10:14, 28 July 2006

Problem

Solution

See also