You are not logged in.
Pages: 1
In square ABCD, E is the midpoint of BC, and F is the midpoint of CD. Let G be the intersection of AE and BF. Prove that DG = AB.
Could someone explain how to prove DG=AB?
Offline
hi harrychess,
If you rotate BFC 90 degrees around the centre of the square, you get AEB. So this proves that AE is perpendicular to BF.
Rotate AEB similarly and mark H on AE, the equivalent point to G.
Show that triangles BFC, BEG and DGH are similar.
Call the length of a side of the square 2a.
Using the similar triangles, find in terms of a, the following:
BE; BF; GE; AH; HG; DG.
This last will give you the required result.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
Offline
Hi bob. I don't get what you mean by mark H on AE and how we prove that DGH is similar to BFG and BEG
Thanks
Offline
Hopefully, this diagram will help.
I seem to have miss out a label on a point. Call the intersection of AB and DH, J.
I rotated BCF 90 to get ABE and then once more to give DAJ.
angle FBC is common to triangles FBC and BGE and BGE = BCF = 90.
Triangles AHJ and AGB are similar with AH = half AG.
So DH splits ADG into two congruent triangles, so angle HGD = ADH = FBC.
So the three triangles have the same angles and so are similar.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
Offline
Pages: 1