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Medians
and of are perpendicular at point . Prove that $AB = CG$.In your diagram,
should appear to be a right angle.Please try to explain using geometry only terms, like try not to use trigonometry or anything above geometry.
Last edited by Enshrouded_ (2015-08-01 13:40:42)
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hi Enshrouded_
I think this works:
As AGB = 90, then a circle centred on the midpoint, D, of AB will go through G.
(EDIT: As I read this through, I see I haven't used the circle at all. Oh well, not to worry.)
Extend AY to E, where AG = GE, and BX to F, where BG = GF.
Join CE and CF.
As G is the median BG = 2GX => GX = XF.
Triangles AGX and CFX are congruent, (SAS) so FC = AG.
Similarly CE = BG.
So triangles AGB, CFG and CEG are congruent (SSS)
So CG = AB.
Bob
Last edited by Bob (2015-08-01 19:56:13)
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