Question In Space Geometry

  • Thread starter xenogizmo
  • Start date
  • #1
30
0
Hey everybody..
I got this space geometry question that had me stumped.. Please help me out.. Here goes:

ABCD is a parallelogram. The plane π was drawn passing through the diagonal AC.
If DE was perpendicular to π, BF was perpendicular to π.
Prove that BF = DE

please see the attached drawing
 

Attachments

Answers and Replies

  • #2
120
0
are E and F on π?
 
  • #3
30
0
yes, I guess they are.. since ED and BF are both perpendicular on π :confused: any help??
 
  • #4
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
19
Well, what have you tried?
 
  • #5
30
0
Hey I think I got it!! :eek:
First draw the diagonal BD


DE and BF are both perpendicular to π
so DE and BF are parallel

This means that the angles BDE and FBD are equal.

And since ABCD is a parellogram, then AB and CD are parrallel.
so the angles BDC and DBA are equal

We then conclude that the angles CDE ad ABF are equal
(because we already proved that the bisects make equal angles)

in the triangles DCE and ABF
angle E=F=90 degrees
D = B proven
C = A 180-both angles

so the triangles are congruent

this means: ED/BF = CE/AF = CD/AB

and since ABCD is a parallelogram, then CD = AB

so DE/BF = CD/AB = 1/1

that concludes to DE = BF :cool:

Am I right here guys? :confused:
Sorry for the crappy solution but I know all the theorems in arab so it's hard for me to write in english cuz I dont knwo the exact terms. :biggrin:
 

Attachments

Last edited:
  • #6
30
0
:cool: I guess no newz is good news.. finally I can get this question off my back :rolleyes:
 
  • #7
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,051
18
I'm not convinced that you've got it right. It's not clear how you show <CDE = <ABF. Sorry to put the question back on your back after all this time.
 
  • #8
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,051
18
Try to look at the triangles DEM and BFM where M is the midpoint of AC.
You've shown that <BDE = <DBF.
And since M lies on AC, EM and FM are lines in the plane. Hence, <DEM =<BFM = 90.
Finally, DM = BM = 1/2(BD) since the diagonals intersect at their midpoints.
Thus, the triangles DEM and BFM are congruent; so DE = BF.

'Congruent' means identical in shape and size, while the term 'similar' is used for objects that are identical in shape. So, when 2 triangles have the same angles, they are similar, but if they also have an equal side, they become congruent.

Hope this helps.
 
  • #9
robphy
Science Advisor
Homework Helper
Insights Author
Gold Member
5,757
1,052
I think this fills in the gap to first show that triangles ABF and CDE are similar.

Since AB || CD, and is each cut by AC, then <BAC = <DCA.
(Assuming that F and E are on AC, then <BAF=<DCE).

As noted by xenogizmo, <E=<F.

So, <ABF=<CDE.

Thus, triangle ABF is similar to triangle CDE.
 

Related Threads on Question In Space Geometry

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
1
Views
3K
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
5
Views
2K
Top