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2006 Romanian NMO Problems/Grade 9/Problem 2

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Problem

Let $\displaystyle ABC$ and $\displaystyle DBC$ be isosceles triangles 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 $\displaystyle DP \perp EF$ .

Solution

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See also

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2006_Romanian_NMO_Problems/Grade_9/Problem_2"

Categories: Problems with no solution | Olympiad Geometry Problems

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