AoPS - Problem

Let $M$ be the incenter of the tangential quadrilateral $A_1A_2A_3A_4$. Let line $g_1$ through $A_1$ be perpendicular to $A_1M$; define $g_2,g_3$ and $g_4$ similarly. The lines $g_1,g_2,g_3$ and $g_4$ define another quadrilateral $B_1B_2B_3B_4$ having $B_1$ be the intersection of $g_1$ and $g_2$; similarly $B_2,B_3$ and $B_4$ are intersections of $g_2$ and $g_3$, $g_3$ and $g_4$, resp. $g_4$ and $g_1$. Prove that the diagonals of quadrilateral $B_1B_2B_3B_4$ intersect in point $M$. [asy][asy] import graph; size(15cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-9.773972777861085,xmax=12.231603726660566,ymin=-3.9255487671791487,ymax=7.37238601960895; pair M=(2.,2.), A_4=(-1.6391623316400197,1.2875505916864178), A_1=(3.068893183992864,-0.5728665455336459), A_2=(4.30385937824148,2.2922812065339455), A_3=(2.221541124684679,4.978916319940133), B_4=(-0.9482172571022687,-2.24176848577888), B_1=(4.5873184669543345,0.057960746374459436), B_2=(3.9796042717514277,4.848169684238838), B_3=(-2.4295496490492385,5.324816563638236); draw(circle(M,2.),linewidth(0.8)); draw(A_4--A_1,linewidth(0.8)); draw(A_1--A_2,linewidth(0.8)); draw(A_2--A_3,linewidth(0.8)); draw(A_3--A_4,linewidth(0.8)); draw(M--A_3,linewidth(0.8)+dotted); draw(M--A_2,linewidth(0.8)+dotted); draw(M--A_1,linewidth(0.8)+dotted); draw(M--A_4,linewidth(0.8)+dotted); draw((xmin,-0.07436970390935019*xmin+5.144131675605378)--(xmax,-0.07436970390935019*xmax+5.144131675605378),linewidth(0.8)); draw((xmin,-7.882338401302275*xmin+36.2167572574517)--(xmax,-7.882338401302275*xmax+36.2167572574517),linewidth(0.8)); draw((xmin,0.4154483588930812*xmin-1.847833182441644)--(xmax,0.4154483588930812*xmax-1.847833182441644),linewidth(0.8)); draw((xmin,-5.107958950031516*xmin-7.085223310768749)--(xmax,-5.107958950031516*xmax-7.085223310768749),linewidth(0.8)); dot(M,linewidth(3.pt)+ds); label("$M$",(2.0593440948136896,2.0872038897020024),NE*lsf); dot(A_4,linewidth(3.pt)+ds); label("$A_4$",(-2.6355449660387147,1.085078446888477),NE*lsf); dot(A_1,linewidth(3.pt)+ds); label("$A_1$",(3.1575637581709772,-1.2486383377457595),NE*lsf); dot(A_2,linewidth(3.pt)+ds); label("$A_2$",(4.502882845783654,2.30684782237346),NE*lsf); dot(A_3,linewidth(3.pt)+ds); label("$A_3$",(2.169166061149418,5.203402184478307),NE*lsf); label("$g_3$",(-9.691606303109287,5.354407388189934),NE*lsf); label("$g_2$",(3.0889250292111465,6.727181967386543),NE*lsf); label("$g_1$",(-4.763345563793459,-3.4725331560442676),NE*lsf); label("$g_4$",(-2.663000457622647,6.878187171098171),NE*lsf); dot(B_4,linewidth(3.pt)+ds); label("$B_4$",(-1.5647807942653595,-3.0332452907013523),NE*lsf); dot(B_1,linewidth(3.pt)+ds); label("$B_1$",(4.955898456918535,-0.6583452686912173),NE*lsf); dot(B_2,linewidth(3.pt)+ds); label("$B_2$",(4.104778217816637,5.0661247265586455),NE*lsf); dot(B_3,linewidth(3.pt)+ds); label("$B_3$",(-3.4454819677647146,5.656417795613188),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

Let $\Gamma$ be the incircle of $A_1A_2A_3A_4$. All the following indexes are in modulo $4$. Let $T_i$ be the tangency point of $A_iA_{i+1}$ with $\Gamma$ and let $A'_i$ be the inverse of $A_i$ wrt $\Gamma$, thus $M,A'_i,A_i$ are collinear and $A'_i$ is the midpoint of $T_iT_{i-1}$. Also $B_iB_{i+1}\perp MA'_{i+1}\Rightarrow A_{i+1}$ is the pole of $B_iB_{i+1}$. By La Hire's Theorem we get $A'_iA'_{i+1}$ is the polar of $B_i$. So we want to prove that $A_iA_{i+1}$, $A_{i+2}A_{i+3}$ and the polar of $M$ are concurrent. The inverse of $M$ wrt $\Gamma$ is $P_{\infty}$, so we want to prove that $A_iA_{i+1}\cap A_{i+2}A_{i+3}=P_{\infty}\Leftrightarrow A_iA_{i+1}\parallel A_{i+2}A_{i+3}$, but this is Varignon's Theorem. Q.E.D.