



 
Chris responded with a proof that uses coordinates and then Shamik produced his own proof. We have Shamik's proof first and then Chris'. Please, refer to the figure. Here is the solution. Let Q be the midpoint of DE. Join QA, QM, QN, MD, MC and NE. Triangle (AMN) = Triangle (AMQ) + Triangle (QMN) + Triangle (QNA) Q, M and N are midpoints of DE, BE and CD, respectively. Therefore, QM is parallel to AB and QN is parallel to AC. And, hence, Triangle (AQM) = Triangle (DQM) and Triangle (AQN) = Triangle (EQN) So, Triangle (AMN) = Triangle (AQM) + Triangle (QMN) + Triangle (QNA) = Triangle (DQM) + Triangle (QMN) + Triangle (EQN) Thus, Triangle (AMN) = Quadrilateral (DMNE) ………………. (1) M is the midpoint of BE and so Triangle (DME) = (1/2)* Triangle (BDE) while Triangle (CME) = (1/2)* Triangle (BCE) è Triangle (DME) + Triangle (CME) = (½)*{Triangle (BDE) + Triangle (BCE)) è Quadrilateral (DMCE) = (1/2)*Quadrilateral (DBCE) Similarly, Quadrilateral (DMNE) = (1/2)*{Quadrilateral (DMCE)} = (1/4)*{Quadrilateral (DBCE)} ……. (2) From (1) and (2) we get, Triangle (AMN) = (1/4)*{Quadrilateral (DBCE)} Shamik Chris' proof We simplified the problem a bit by first shrinking the quadrilateral BCED by ½: Take C' and D' the midpoints of BC and BD, then show that (AMN) = Area(BC'MD'). Using the coordinates shown in the figure, we compute and Was your argument this easy? We will let you know if we find a simple argument that avoids coordinates. Chris
 


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