+18 Medians $\overline{AX}$ and $\overline{BY}$ of $\triangle ABC$ are perpendicular at point $G$. Prove that $AB = CG$.
+128460 
Triangle ABC is isosceles with AC = AB
And CD is an altitude....and in an isosceles triangle.....the altitude bisects the base...so
AD = BD
And by SSS....triangle DAC is congruent to triangle DBC
So...angle DCA = angle DCB
And CY = CX
And GC = GC
And by SAS....triangle GCY is congruent to triangle GCX
Angle AGB = 90 = Angle YGX = vertical angles
And since angle CGX = CGY then angle YGX is bisected
So angle CGX = angle CGY = 45
The angle DGA must also =45
And angle GDA = 90....so angle XAB = 45
And we have triangles CGX and AXB
And since angles CXG and BXA are supplemental....their sines are equal
So....by the Law of Sines
sin BXA/ AB = sin XAB / BX
sin BXA / AB = sin 45 / BX
And
sin CXG / CG = sin CGX / CX
sin CXG / CG = sin 45 / CX
But since BX = CX
Then sin 45/ BX = sin45/ CX
Which implies that
sinBXA / AB = sin CXG / CG
CG sin BXA = AB sin CXG
And sin BXA = sin CXG
So
CG = AB
