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

 Jun 9, 2019
 #1
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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

 

 

cool cool cool

 Jun 9, 2019

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