Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 833.

## Wednesday, December 19, 2012

### Problem 833: Parallelogram, Diagonal, Similarity, Triangle, Proportion

Labels:
parallelogram,
proportions,
similarity,
triangle

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Construim paralela prin punctul C la dreapta BE si notam cu M intersectia acestei paralele cu dreapta AD.Va rezulta ca patrulaterul BMNC este paralelogram =>BC=EM=AD.Aplicand teorema lui Thales in triunghiul ACM cu EF paralela cu CM va rezulta AF/FC=AE/EM si in final AF/FC=AE/AD

ReplyDeleteNote that triangles AFE and CFB are similar ( Case AA)

ReplyDeleteso AE/BC=AF/FC

but BC=AD so AE/AD=AF/FC