Side $AB$ of the square $ABCD$ is also the hypotenuse of right triangle $ABP$ ($ABP$ lies outside $ABCD$).Prove that the angle bisector of $\angle APB$ bisects the area of $ABCD$.
My efforts:
I've constructed 4 coungrent circles which have as centers the midpoints of the sides of square $ABCD$ and have as radius the distance which goes from the center of the square to the midpoint of the corresponding side.
Now,since $\Delta ABP$ is a right triangle lying outside $ABCD$ and it has as hypotenuse side $AB$ ,we have that point $P$ must be on the upper half portion of arc $AB$.
Then i construct segment $PO$ where $O$ is the center of square $ABCD$,
now if we consider $\angle APO$ we see that it is equal to $\angle ABO$ (equal inscribed angles) ,and $\angle ABO=\angle OPB$ since they intercept equal segments $AO,BO$ and we know that these segments are equal because they're formed by joining the endpoints of the areas in common by each pair of circles,which are by simmetry the same.
Therefor we see that that segment $PO$ is the angle bisector of $\angle APB$ and that it passes through the center of square $ABCD$ ,and every line passing through the center of a square bisects its area.
Question:
Is this thinking correct ?I know that many parts of my thinking are redundant and could be simplified even more but is it in general okay ?
I would also like to see if there are (i am sure there are) better ways to approach the problem.
P.S: Forgive me if i've made any english mistakes or the phrasing wasn't really clear and feel free to edit in case.
Thanks in advance.
Best Answer
Your approach looks fine to me. Indeed, the most natural way to show that a line bisects the area of a square is to show that it passes through the center of that square. Here this follows from the fact that $O$ is the midpoint of the arc $AB$ of the circumcircle of $\triangle APB$ and the fact that the inner angle bisector in a triangle passes through the midpoint of the opposite arc of the circumcircle.