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# geom. question

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Draw triangle ABC and let D be the midpoint of AB. Let E be the midpoint of CD. Let F be the intersection of AE and BC. Draw DG parallel to EF meeting BC at G. Prove that BG = GF = FC.

Jan 11, 2018
edited by Guest  Jan 11, 2018
edited by Guest  Jan 11, 2018

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See the pic of the situation, here :

Note that in triangle ABF.....that  AF  is a base  and DG is  a segment parallel to this base....but a segment in a triangle that is parallel to a base splits the sides of the triangle proportionally

Therefore  :   BD / DA  =  BG/ GF

But D  is the midpoint of AB...so.... BD  = DA

But....this also means that  BG  =  GF

And  DGC  forms another triangle with DG  as a base  and EF a segment parallel to this base

And for the same reason as above :

DE / EC  =  GF / FC

But  E  is the midpoint of  CD....thus....DE  = EC

But...this also means that GF =  FC

But....it was proved that  BG =  GF

So.....by transitivity ...... FC  also  equals BG

Therefore    BG  = GF  =  FC

Jan 11, 2018
edited by CPhill  Jan 11, 2018
edited by CPhill  Jan 11, 2018