# Prove that this figure is a rhombus

• May 18th 2010, 10:30 PM
Ragnarok
Prove that this figure is a rhombus
I am working through Serge Lang's book on geometry, and I am having trouble for some reason on this proof:

Given that AC is on the perpendicular bisector of BD, and that BD lies on the perpendicular bisector of AC. Prove that quadrilateral ABCD is a rhombus (a rhombus is a parallelogram whose four sides have the same length).

It is easy to prove that the four sides are of equal length, but I don't know how to prove that the pairs of sides are parallel. I think I should prove that the interior angles are supplementary, but I'm not sure how to do that. Any tips?
• May 18th 2010, 10:43 PM
sugarT
one way to do this could be by first siding the sides by Cosine Rule (or Pythagoras) then applying sine rule in one of the traingles to find an accute angle. Find the alternate angle on another traingle, and if they turn out to be equal, then if two angles are alternately equal, then the lines must be parallel.

hope that helps :)

-sugarT

EDIT: easier would be just to do Cosine Rule to find the alternate angles.
• May 18th 2010, 11:43 PM
Ragnarok
Thank you so much for your suggestion! I don't think the book covers the sine/cosine rules though. Is there a simpler way to prove it? It seems like I've done a similar thing before, but I'm blanking out for some reason.
• May 19th 2010, 12:14 AM
sugarT
My pleasure. A simpler method could be by using triangular congruency, two sides and an angle in between. All triangles seem to have corresponding equal sides and an equal angle in between the two sides. Thus, they are congruent and hence we can find alternate equal angles which are in fact corresponding angles of the congruent triangles . Then from that say they must be parallel.
• May 19th 2010, 12:37 AM
priyanka0528
Lets consider ABCD to be a quadrilateral
AC intersects BD perpendicularly at say O
AOD = COB = 90
BO is common
ABO = CBO since when BD intersects AC perpendicularly
hence triangle ABO=triange BOC
hence AB=CB and AO = OC
Similarly it can be proved that triangle ADO = triangle DOC