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In triangle ABC, AB=BC=12. Side AC extended through C a length equal to itself to a point D. Point E is on AB; DE intersects BC at F and BF equal to 8. Find AE ?

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The answer is $AE=BE=6$ because in the triangle ABD, BC is a median and F is the intersection point of medians, therefore DE is also median. Since, $BF=8$ then $CF=4$, but we know that the intersection point of medians splits them into two segments which lenghts have quatient 2:1 from the vertex.

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+1 Nice way of seeing it, in this special setup. You could fill in a bit more details about why F is the centroid (of what triangle). – Calvin Lin Feb 26 '14 at 6:56

BC divides AD in half. So in triangle ABD, BC is the median. BF:FC = 2:1, so F is the centroid. Since ED passes through centroid, it is also a median. Hence AE = BE = 6.

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Hint: Apply Menelaus' theorem to triangle $ABC$ and transversal $DEF$.

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Yes, more "classical" theorems that don't get much "air play" nowadays... – RecklessReckoner Feb 26 '14 at 7:16
@RecklessReckoner If you are a student training for olympiads, they are important. But once you are out of that phrase of your life, very few people care about Euclidean Geometry anymore. – Calvin Lin Feb 26 '14 at 8:05

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