On the sides $BC$ and $CD$ of the square $ABCD$, the points $M$ and $N$ are selected in such a way that $\angle MAN= 45^o$. Using the segment $MN$, as the diameter, we constructed a circle $w$, which intersects the segments $AM$ and $AN$ at points $P$ and $Q$, respectively. Prove that the points $B, P$ and $Q$ lie on the same line.
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Tags: geometry, square, collinear
i3435
08.09.2020 19:46
Can someone provide a diagram, my diagram doesn't seem to be right.
Quentissential
08.09.2020 20:30
[asy][asy] pair A=(0, 0); pair B=(0, 10); pair C=(10, 10); pair D=(10, 0); pair EE=(10, 5); pair F=(5, 10); draw(A--B); draw(B--C); draw(C--D); draw(A--D); draw(A--EE); draw(F--A); draw(F--EE); [/asy][/asy] is the start of a diagram
natmath
08.09.2020 20:44
By symmetry, if $B,P,Q$ are collinear then so are $P,Q,D$, which means we need to prove that $PQ$ lie on the diagonal.
vanstraelen
08.09.2020 21:36
Given $A(0,0),B(a,0),C(a,a),D(0,a)$. Choose the points $M(a,\lambda)$ and $N(\frac{a^{2}-a\lambda}{a+\lambda},a)$, so that $\angle MAN= 45^{\circ}$. Equation of the circle $x^{2}-\frac{2a^{2}}{a+\lambda} \cdot x + y^{2}-(a+\lambda)y+\frac{a(a^{2}+\lambda^{2})}{a+\lambda}=0$. Points $P(\frac{a^{2}}{a+\lambda},\frac{a\lambda}{a+l})$ and $Q(\frac{a-\lambda}{2},\frac{a+\lambda}{2})$. Equation of the line $PQ\ :\ y=a-x$.
rafaello
08.09.2020 22:24
It is well-known that $A$ is the excenter of triangle $CNM$ (can be shown by phantom points for example). Thus, $$\angle AMB=\angle AMN=\angle PMN=180^{\circ}-\angle PQN=\angle AQP=\angle AQB.$$Hence, $AQMB$ is cyclic and $\angle QBM=\angle QAM=\angle MAN =45^{\circ}$. Similarly, $\angle PDN=\angle MAN =45^{\circ}$. Hence $B,P,Q,D$ are collinear and we are done.