Question 129774
If you construct the altitude you will see that the sides {{{AB}}} and {{{CB}}} are {{{not}}}{{{ congruent}}}.
 
I think the way to prove {{{opposite}}} will be best way to shove that altitude {{{BD}}} would {{{not}}} bisect {{{AC}}}.

So, let's assume that the altitude {{{does}}} bisect {{{AC}}}.
 
Then {{{AD}}} and {{{DC}}} are {{{congruent}}} (by the def. of bisector),

 so triangles {{{BAD}}} and {{{BDC}}} are congruent (by the side-angle-side theorem)
 if {{{BA}}} and {{{BC}}} are {{{congruent}}}, then {{{ACB}}} {{{is}}}{{{ isosceles}}}
 
Hence, if the {{{altitude}}} bisects {{{AC}}}, then the triangle {{{must}}}{{{ have }}}{{{been }}}{{{isosceles}}} to start with.