The point $M$ inside a convex quadrilateral $ABCD$ is equidistant from the lines $AB$ and $CD$ and is equidistant from the lines $BC$ and $AD$. The area of $ABCD$ occurred to be equal to $MA\cdot MC +MB \cdot MD$. Prove that the quadrilateral $ABCD$ is a) tangential (circumscribed), b) cyclic (inscribed). (Nairi Sedrakyan)
Problem
Source: Tournament of Towns, Senior A-Level , Fall 2019 p5
Tags: tangential, cyclic quadrilateral, Cyclic, geometry, area, Kvant
VIP_LION_11
25.04.2020 04:19
B) Let's for definiteness consider the angle bisector formed by straights $AD$ and $BC$ ( or axis of symmetry corresponding band) and reflect relatively her both A and D .Let points A' and D' corresponding forms .Notice that for areas we have true equality $ area[AMD]+area[BMC]=area[A'MD']+area[BMC] = area[A'MC]+area[BMD'] $ and there we note that doubled area of a triangle $BMD'$ doesn't exceed of product length of segments $BM$ and $MD'$ (which is equal to $MD$) then we conclude that in fact, the fact that $M$ belongs to the corresponding bisector leads to an estimate 2×$area[AMD]$+2×$area[BMC]$<= $AM$×$MC$+$BM$×$MD$ with equality $angleA'MC$=90° AND $angle BMD'$=90° analogously we define that belonging to the 2nd bisector 2×$area[AMB]$+2×$area[CMD]$<= $AM$×$MC$+$BM$×$MD$ ADDING UP THIS INEQUALITIES WE GET $area[ABCD]$ <= $AM$×$MC$+$BM$×$MD$ but in Conditions in inequality we have equality then we get $angleA'MC$=$angleBMD'$=90° then we have $angleMAD'$+$angleMCB$=90°= $angleMDA$+$angleMBC$ conducting similar reasoning for axis of symmetry or bands between lines $AB$=$DC$ we will get new equalities $angleMAB$+$angleMCD$=90° =$angle MBA$+$angleMDC$ then opposite angles of our quadrilateral in sum gives us 180° A) sum of angles of quadrilateral under which from the point $M$ to the opposite sides of the quadrilateral are visible $ABCD$ EQUAL to 180 ° it gives us point availability $M$ isogonal conjugate of $ABCD$ but we know that $M$ lies on angle bisectors formed by opposing sides but $M$ is should be isogonally conjugate to self that gives us point $M$ is the intersection point of internal bisectors of quadrilateral $ABCD$ (Sorry for my English)