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

charluu Jun 9, 2019

#1**+2 **

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

CPhill Jun 9, 2019