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

#1**+3 **

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

Add DG to everything

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