They meet at right angles
The intersection of the bisectors of the angles. (Sorry _ my original answer, the bisectors of the sides, was clearly wrong.)
4 sides are equal Diagonals are equal and Perp Bisectors Angles are right Consecutive and diagonal angles add up to 180.
The bisectors of the angles of a triangle meet at the incentre.
Congruent (APEX) :P
They meet at right angles
The diagonals of a square are perpendicular (they intersect and form right angles). But they are angles bisectors since they bisect each pair of opposite angles. A perpendicular bisector actually bisects a side of a figure.
It's fairly trivial to prove that the angles formed by the angle bisectors of any rhombus (including squares) are right angles.
Yes.
I am working on the same exact proof right now and i am lost
yes
The intersection of the bisectors of the angles. (Sorry _ my original answer, the bisectors of the sides, was clearly wrong.)
4 sides are equal Diagonals are equal and Perp Bisectors Angles are right Consecutive and diagonal angles add up to 180.
I expect "consecutive angles" are any pair that aren't opposite. Since they are co-interior angles between parallel lines, they are supplementarty (i.e. total 180 deg). When you bisect them, the bisectors join to form a triangle. Two of its angles are halves of the "consecutive angles", and so they total half of 180 deg, which is 90 deg. Hence the third angle is 90 deg (to give angle sum of the triangle as 180 deg), so the bisectors are perpendicular.
The bisectors of the angles of a triangle meet at the incentre.
The bisectors of the angles of a triangle are concurrent at a point called the incentre which is also the centre of the inscribed circle that touches all three sides.
Yes.