Doesn't make sense. You cannot bisect a vertex!

But the perpendicular bisectors of the 3 sides of any proper triangle on a plane are concurrent. For this specific case, if P, Q and R are the mid-points of the 3 sides BC, CA and AB: you can find their mid-points P, Q and R, the gradients of BC, CA and AB, and therefore of their perpendiculars. Then you can find the equations of these 3 perpendiculars. Find where 2 perps intersect; show this point of intersection lies on the 3rd perpendicular.

## Bookmarks