please help me with this problem!! thank you:

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.

Guest Jan 11, 2018

edited by
Guest
Jan 11, 2018

edited by Guest Jan 11, 2018

edited by Guest Jan 11, 2018

#1**+2 **

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

CPhill Jan 11, 2018