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Mathematics 14 Online
OpenStudy (anonymous):

Proofs?

OpenStudy (anonymous):

What do you mean ma'am?

OpenStudy (anonymous):

Given: CD ⊥AE and CD GH || ; Prove: GH ⊥AF .

OpenStudy (jack1):

medal for "ma"am"

OpenStudy (anonymous):

Thank you Sir!

OpenStudy (jack1):

"Given: CD ⊥AE and CD GH || ; Prove: GH ⊥AF" is this meant to be : Prove: GH ⊥AE" because we haven't mentioned point F before the end of the Q....? so: just to be clear: CD is perpendicular to AE (at 90 degrees to) and CD runs parallel to GH |dw:1387931678151:dw|

OpenStudy (anonymous):

here is the chart i don't know how to make it into a photo

OpenStudy (jack1):

so in the above pic, CD and GH are parallel lines (therefore at 0 degrees difference to eachother) and AE is perpendicular to CD (therefore at 90 degrees difference to it) therefore AE must also be perpendicular to GH (as CD and GH are parallel lines, so AE must intersect at 90 degrees to both)

OpenStudy (anonymous):

thanks alot!!!!

OpenStudy (jack1):

ah, ok, so AF is an extention of straight line AE so above statement still holds true and your welcome hey, gud luck an happy chrissie ;D

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