\(\frac{a+b}{a-b}=\frac{a+c}{c-a}\)
=> ( a + b ) ( c -a ) = ( a + c ) ( a - b )
=> a ( c - a ) + b ( c - a ) = a ( a - b ) + c ( a - b )
=> ac - aa + bc - ba = aa - ab + ca - bc
=> - aa - aa = - bc - bc
=> - 2 a 2 = - 2 bc
=> a 2 = bc
Vậy \(\frac{a+b}{a-b}=\frac{a+c}{c-a}\)thì a 2 = bc