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.
+128407 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
