GSPk{ F1?capmdthm AWVCZD t{gl AV@D@ZD tr wxT7 AUxX?xH?x?x?x 7x7x|7xl7x\7xL7x7x7x7x7x7xt 7xL 7xCD t{ 7x ATX?x@C\D`4DPD 7xC\D7x@DD7xC$\D@DD@DD tz /To trisect the area of any given rectangle ABCDt   F  Pȸ@ZD=Ci GLDC t/4D~1\ B\SKETCH\SORTED~1\2D-GEO~1\TRIAN~1\Trisec~1.gspcDC t/4 AdeGTH8$GCCt There are two theories present here: The first is applicable when both X and Y lie on the line CD - here the rectangle ABCD is trisected by drawing lines PX and PY through M1 and M2, respectively. Where M1 and M2 are the midpoints of two lines that are parallel to sides AD and BC and that trisect the area ABCD. The second is applicable when either X or Y no longer lie on the line CD. The problem then becomes one of bisecting the quadrilaterals PYDA and PXCB respectively. Click on "Show Bisect Area Figure" Button below.tsl Drag Verticiest1U m!07 q@,#H7 jTo bisect any irregular quadrilateral figure by a line drawn through one of its angular points (verticies)tP&    dPROOF. Let PXCB be the given figure, and P the given point. Join XB, and bisect it in E. Then a figure PECX would halve the quadrilateral, for PEB = PBX/2, and BEC = XBC/2. Join PC, and through E draw EY parallel to PC to meet CD in Y. Join PF, and the figure will be bisected since PYC =PEC (Euclid i. 37) (Euclid i. 37: Triangles on the same base, and between the same parallels, are equal in area)tr wC X'?'?o8j~q ƝtlACD t{ P C @c  N"o  Zȣ| #  PR@DD t{glC B00S4j3j/Os@D@ZD thmC P,ۄ*(>LL:EN@ۄm(( VVCZDtrw AOVC]D t{qv]# AP@D(]#D`DCD]CDdC@D]#D`DCD@D\D t}qvD AQCFD]CD#D`DDjDdC@D]#DjD#D`DdCC`D t|qvd BE AR, and bisect it in E. Then a igure AECB would halve the quadrilaterD`D t"H8 Hide t N  Show t amw Move B->AV t Jq` Move P->AW t 2kH Move X->AU8?x?x4?x|?x?x ?xP@8 /x/x/x x/x t k0 Move C->AT7x@7xl7x7x 7x?xH?x?x?x?x?x x ?x t.4M@ n0LL:EN@ۄm(( V  t44J Show Proof '?'?o8j~q ƝtlA  tqfo@@@pppppppppppppppppppppppppppppppppppppp@D@ZDCD?  tg@@npppppppp@@@@VCZD@DD? tgwppppm@@@@@ppppppppppppppppCDVCZD? tq +Cl Sg-hBCCd ЂCuCD | "DD=CP -@DDCD?  tzf7 ykCJDCD@"CP (\ ED (7 CC@D@ZD@DD? tfmCDjjD.CwCC(%'>CC'>CCCC CLC HVCZD@D@ZD?t's= Move P->AO t>rT Move B->AP tUqk Move X->AQ  tlr Move C->ARect it in E. Then a igure PECB would halve the quadrilater BDC/2.t Reset VerticiesX(>dL:EN,UmwX((wX  t~/4 $>  P?" +1: +?:袜GX9CCt). mxO00O0O0O0O0O0O0XOO0O0xO0hO0 .....cDcDD?t/4}7?  Cd4C\ xec~1.gspew ` f7DC t9>Fpppppppppppppppp@C`D t-C Move point P along side AB +?:袜G**t~&+δ /4 AM'X9C.C *tɿ D?" +7T$9HK? VCf s1D)CC:@ZDBC#Di@iCfLL:EN@ۄm(( V@DǗD 1#t#h(mppppG@@@@@@@@@@@@@@@@p@pppppppppppppppbCtZD $1 tw(:! p12F%5?F%5?X9CC*CC2t=B  GhBsi>CDZlt] m1tch01.gsp Area PBY = Area(Polygon PBY) = ?t9> E @c  N"o  Bȣ| #  PR?C`D Atv Hide Figure*wX(>dL:EN,UmwX((wX  `xp?x)  !"#$9:>?@ADEFt Show Bisect Area Figure>dL:EN,UmwX((wX  |x?x )  !"#$9:>?@ADEF t}. apX9CCcDfOvD 6B tC*1 pACDh c@ g>lC cDf:9$#"!  t| t01?@ZDBBi@~;AvDcDf>`#BX#&Cf UQDD ct) arCCT~Df