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.
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