look at the figure !
ABCD is parallelogram. Find ED ?
BC = (5/3)x because BFC and EFD are similar triangles
Do you know how to find BD?
how do u know BFC and EFD are similar ?
Because the three triangles and clipped triangles are enclosed in a parallelogram, angle FBC = angle EDF and angle EFD = angle BFC. Right?
The second part isn't because of the parallelogram, just because they are opposite.
oh ok, i see thats right... so, next ?
I had something in mind with BD but lost it. Can you think how BD might help?
hmmm, for this time no idea...
if only we had some angles... i'm going to see if i can find anything about parallelograms on the internet. geometry was a long time ago for me.
"The sum of the distances from any interior point of a parallelogram to the sides is independent of the location of the point." we should be able to use that methinks
did you study Viviani's theorem?
i think we can create an equation using the similar sides FD and FB with Viviani's theorem.
but maybe not
the only thing i'm mostly sure is useful is the similar triangles. without more info, i'm kindof lost.
yea, i think like that... but still stuck Viviani's theorem? i dont know it
if you haven't covered that (i just now learned of it) it's probably not relevant.
ok, nope. thanks u have remember to me about similarity
triangles with 2 of 3 identical angles (the third will always be the same if the other two are) will be proportional.
you good on why that applies to EFD and BFC?
that gets us a relationship between x and the long sides' length. gotta be good for something.
is the problem not complete information ?
is that an option?
nothink
you dont think so?
it is a essay question.... i was lost 3 hour to do it :(
wow. that is unfortunate. what class?
grade 9th
without any angles to do some trig, i cant think of anything sorry. i'm gonna get out of here. good luck.
ok, thank you so much....
i would think there has to be something in your notes that applies to this. check there. you're welcome.
yups...
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