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$$ABCD - square $$ $$M\in AB, N \in AD, P \in DC : BM = AN = DP$$ 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!

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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.

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  • $\begingroup$ @liambro, Thank you for your response! I got that $\angle MNP = 90^\circ$, and DPBM is a parallelogram, but I don't understand why we can state what the problem wants from here. $\endgroup$ – Nikol Dimitrova Apr 16 at 17:55
  • $\begingroup$ We have that $O \in MP$, OP = OM but I don't understand the part of your solution with DPBM. What's the point to use it? $\endgroup$ – Nikol Dimitrova Apr 16 at 17:59
  • $\begingroup$ We want to prove that the midpoint of MP is the intersection of diagonals of the square. The point of intersection of the diagonals of ABCD is the midpoint of DB. How to prove that midpoint of DB is the midpoint of MP? That statement is equivalent to DPBM being a parallelogram. $\endgroup$ – liaombro Apr 16 at 18:31
  • $\begingroup$ @liambro, I think I got it. Can you tell me if I'm right? $\triangle MNP$ is right-angled, thus the circumcentre is the midpoint of the hypotenuse: $O \in MP, OM = OP$. DPBM is a parallelogram (DP || MB, DP = MB) and O is the midpoint of PM, thus DB passes through O: OD = OB => AC passes through O. $\endgroup$ – Nikol Dimitrova Apr 16 at 18:52
  • $\begingroup$ @liambro, If I understand the key here is that we have the midpoint O of PM - diagonal in the parallelogram; thus the other diagonal - DB, passes through this midpoint and DO = OB. Now, O is the midpoint of DB; therefore the other diagonal of ABCD - AC, passes through O: AO = OC. $\endgroup$ – Nikol Dimitrova Apr 16 at 19:09
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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.

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$\bigtriangleup DPO \cong \bigtriangleup OMB$ 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.

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