Let vertices of an acute-angled triangle are `A(z_1),B(z_2),a n dC(z_3)dot` If the origin `O` is he orthocentre of the triangle, then prove that `z_1( z )_2+( z )_1z_2=_2( z )_3+( z )_2z_3=z_3( z )_1+( z )_(3)z_1dot`

Explanation

Here, O is orthocenter, then
`OA_|_BC" "(becauseAD_|_BC)`
`:." "arg((z_(1)-0)/(z_(2)-z_(3)))=(pi)/(2)`
`:.(z_(1)-0)/(z_(2)-z_(3))" is purely imaginary."`
`implies(z_(1)-0)/(z_(2)-z_(3))+bar(((z_(1)-0)/(z_(2)-z_(3))))=0`
`implies(z_(1))/(z_(2)-z_(3))+(bar(z)_(1))/(bar(z)_(2)-bar(z)_(3))=0`
`impliesz_(1)(bar(z)_(2)-bar(z)_(3))+bar(z)_(1)(z_(2)-z_(3))=0`
`impliesz_(1)bar(z_(2))+bar(z)_(1)z_(2)=z_(1)bar(z)_(3)+bar(z)_(1)z_(3)" "(1)`
Similarly, `OB_|_AC`
`impliesz_(1)bar(z)_(2)+bar(z)_(1)z_(2)=z_(2)bar(z)_(3)+bar(z)_(2)z_(3)`
From, (1) and (2),`z_(1)bar(z)_(2)+bar(z)_(1)z_(2)=z_(2)bar(z)_(3)+bar(z)_(2)z_(3)=z_(1)bar(z)_(3)+bar(z)_(1)z_(3)`