2006 Romanian NMO Problems/Grade 9/Problem 2
Since is isosceles, . Since , , which means that , too. Thus , so by ASA, . This means that . Since , , and , by SAS, , so . Since , $\anlge EDA = \angle BDA - \angle BDE = \angle ADC - \angle MDP = \angle CDP$ (Error compiling LaTeX. ! Undefined control sequence.). Thus . is the angle bisector of , and . This means that . , , and , so by SSS, . Thus and . , so by AA, . Thus . Also, . and , so . By SAS similarity, . is a median and an angle bisector of . Now assume that P' is the point such that DP' is a median of (it is on ). It is on DP, the angle bisector, and since , , but we also showed that . Thus . Since P and P' are on the same ray (), P = P' and P is the midpoint of .