a special case of a widely known fact... so strange for 239mo

What's the widely known fact, can you give any reference?

My bad, I wanted to write this under J4... For problem J4 the fact sounds like this: ABCD is a quadrilateral inscribed in a circle with center O. Then O lies on the Newton-Gauss line of the quadrilateral formed by the bisectors of the angles ABCD

Very beautiful problem!

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Let $AB, BC, CD, DA$ touch the incircle $\odot {I}$ at $P, Q, R, S$ respectively and construct circles $\odot (A, AP), \odot (B, BQ), \odot (C, CR), \odot (D, DS)$. Let $AD, BC$ intersect at ${E}$, $AB, CD$ intersect at ${F}$, and $AC, BD$ intersect at ${G}$. By the lengths condition, we can construct circles $c_1, c_2$ with center ${X}$ and $c_3, c_4$ with center ${Y}$ such that $c_1, c_2$ are externally tangent to $\odot (A, AP), \odot (C, CR)$ and $\odot (B, BQ), \odot (D, DS)$ respectively, and $c_3, c_4$ are internally tangent to $\odot (A, AP), \odot (C, CR)$ and $\odot (B, BQ), \odot (D, DS)$ respectively. Claim: $c_1=c_2$ and $c_3=c_4$. Notice that $\odot (A, AP), \odot (B, BQ), \odot (C, CR), \odot (D, DS)$ are both orthogonal to $\odot {I}$, and that $X, Y, I$ are collinear, the inverstion wrt $\odot {I}$ must send $c_1$ to $c_3$ (since a line cuts a conic at most twice). Similarly, the inverstion wrt $\odot {I}$ sends $c_2$ to $c_4$. The centers of $c_1, c_2$ and those of $c_3, c_4$ are the same respectively, so their radii are also the same, that is $r_1=r_2$ and $r_3=r_4$. Now the magic begins. Call the plane $ABC$ $\alpha$. Create $AA’=AP, BB’=BQ, CC’=CR, DD’=DS, XX’=r_1, YY’=r_3$ in space such that these segments are perpendicular to $\alpha$ and that $X’$ and the other five new points lie on different sides of $\alpha$. Call the cones with vertex $X’, Y’$, axis $XX’, YY’$ and leaning angle $\frac{\pi}{4}$ $Co_1$ and $Co_2$. Obviously $A’, B’, C’, D’$ lie on both $Co_1$ and $Co_2$. Since the intersection of $Co_1$ and $Co_2$.is a conic on a plane plus something at infinity, so $A’, B’, C’, D’$ are on a plane(call it $\beta$). Because the intersection of $PR, AD$ and that of $PR, BC$ is on $\beta$, if they aren’t the same, then $PR \subset \beta$. Similarly, $QS \subset \beta$, meaning that $\alpha \cap \beta$ has two distinct lines $PR, QS$, contradiction! Thus $PRE, QSF$ are collinear respectively and $\alpha \cap \beta$ is exactly line $EF$. $EF$ is the polar of ${G}$ wrt $\odot {I}$, so $IG \perp EF$. However, $Co_1$ and $Co_2$ are actually congruent, so by symmetry $X’Y’ \perp \alpha \cap \beta=EF$. By orthogonal projection $XY \perp EF$, and we get $X, Y, I, G$ are collinear.