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# Special Parts of a Triangle, Continued!

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Medians $$\overline{AX}$$ and $$\overline{BY}$$ of $$\triangle{ABC}$$ are perpendicular at point $$G$$. Prove that $$AB=CG$$.

In your diagram, $$\angle{AGB}$$ should appear to be a right angle.

benjamingu22  Sep 30, 2017
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Here's an  image :

Bisect angle ACB.....and this will form altitude CE

And AC = BC    And angle ACG  = angle BCG    And  CE is common

So by SAS triangle ACG is congruent to triangle BCG

Then AG = BG  so triangle AGB is isosceles....and because AGB is right.......then angles ABG and BAG  =  45°

And since CEB is right and ABG = 45°....then angle BGE is also = 45°

Then triangle  BGE is also isosceles  with EB = EG

Draw  XY......and by hypotenuse-leg, triangle AYG is congruent to triangle BXG

So......AY  =  BX.....so  XY is parallel to AB

And because BX splits BC equally.......then  CD  =  ED

And triangle CDX  is similar to triangle CEB

And  since CD  is (1/2)  of CE then   DX =  (1/2)EB =  (1/2) EG

And angle XAB = angle YXG  =  45°  and since GDX is right, then angle XGD = 45°

So triangles  GDX and  GEB   are similar

And because triangle GDX is isosceles.....then DX = DG

Thus   DX  = (1/2)EG   and by substitution  DG = (1/2) EG  ⇒ 2DG = EG = EB

So

EB + DG  =  3DG

ED  =  3DG = CD

ED + DG  = 4DG  = CD + DG

4DG  =  2 EG  =  CG

But  EG =  EB   ....so....

2EB  = CG

But 2EB  = AB

So....  AB =  CG

CPhill  Oct 1, 2017
edited by CPhill  Oct 1, 2017

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