Geometry/Complex Numbers (1 Viewer)

[ND]

Ok got it, thanks.

 

[riVa0o]

Originally posted by nike33
ok... DZ is parrallel to BE

hence a perpindicular from DZ will perpindiculary cut BE
now as DZ = 2OE (proven using similar triangle rations)
the midpoint of DZ ie m will pass though O in this example (due to the paralled lines)

it has to cut 'o' as MZ = OE = DM and ZE = MO
and, if it didnt cut o then DO =! OC

this is probably a bit old seeing as everyone done it, but can someone explain to me why you can't practically finish the question off once you find that triangle DZC|||TriangleEOC?
because the sides are in the ratio of 2:1, so wouldn't it follow that OC=2DC and therefore DO=OC??

 

[nike33]

Originally posted by riVa0o
this is probably a bit old seeing as everyone done it, but can someone explain to me why you can't practically finish the question off once you find that triangle DZC|||TriangleEOC?
because the sides are in the ratio of 2:1, so wouldn't it follow that OC=2DC and therefore DO=OC??

you can, but as it was a show question...hehe

 

[riVa0o]

does that mean we could just write that ratio thing in an exam rather than all that extra stuff ><

 

[nike33]

maybe, in an exam personally i would do the other stuff, just from experience getting screwed so many 1/2 marks due to picky teachers