Square with circle

Question

$$ABCD - square $$ $$M\in AB, N \in AD, P \in DC : BM = AN = DP$$ I have to show that the intersect of the diagonals is the centre of the circumscribed circle around NMP.

I have noticed that NMP is isosceles, but I can't continue. I would be very grateful if you can help me!

Answer 1

Firstly, $\angle MNP=90^\circ$, since triangles $MAN$ and $NDP$ are equal ($\angle MNA + \angle DNP = \angle MNA + \angle AMN = 90^\circ$)

Secondly, the circumcentre of a right-angled triangle is the midpoint of the hypotenuse.

Thirdly, DPBM is a parallelogram, so the point of intersection of its diagonals bisects them.

Answer 2

Hint: since $\triangle{NMP}$ is isosceles, $NO$ is a height. But $NO=OM$ so $\angle{PNM}$ is a right angle and $O$ is the center of the circle.

To prove that $O$ is also intersection of diagonals, draw two parallel lines thru $O$, one is parallel to $AB$, the other one is parallel to $BC$ and show that these lines will bisect sides of the square.

Answer 3

$\bigtriangleup DPO \cong \bigtriangleup OMB$ as they are similar triangle and also DP = MB. So DO = OB, So O is the midpoint of DB and also PM (and also AC). Now between $\bigtriangleup DPO$ and $\bigtriangleup ANO$ , DP = AN, AO = DO and $\angle PDO = \angle NAO$ So they are congruent. So ON = PO, so ON = PO = OM.