`(ab+ac)/2 = (bc+ba)/3 =(ca+cb)/4=(ab+ac-bc-ba+ca+cb)/(2-3+4)=(2ac)/3`
`⇔(ab+ac)/2 = (bc+ba)/3 =(ca+cb)/4=(-ab-ac+bc+ba+ca+cb)/(-2+3+4)=(2bc)/5`
`⇔(ab+ac)/2 = (bc+ba)/3 =(ca+cb)/4=(ab+ac+bc+ba-ca-cb)/(2+3+4)=(2ab)/1`
`⇔(2ac)/3=(2bc)/5=(2ab)/1`
`⇔(ac)/3=(bc)/5=(ab)/1`
`⇔(ac)/3=(bc)/5`
`⇒a/3=b/5`
`⇔(5a)/3=b (1)`
`⇔(ac)/3=(ab)/1`
`⇔c/3=b (2)`
từ` (1);(2)`
`⇒(5a)/3=b=c/3`
`⇒a/3 =b/5 =c/(15)`