tag:blogger.com,1999:blog-6933544261975483399.post7853387330743952919..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 447: Complete Quadrilateral, Inradii, Exradii, Sides.Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-13328972174274398322010-05-08T00:39:02.752-07:002010-05-08T00:39:02.752-07:00Let external angle of ABCD at A,B,C,D be A',B&...Let external angle of ABCD at A,B,C,D be A',B',C',D'<br />Let O be the excenter of ABE, excircle of ABE touch AB at P, then<br />a/r_a=AP/r_a+PB/r_a=cot(A'/2)+cot(B'/2)<br />similarly c/r_c=cot(C'/2)+cot(D'/2)<br />and a/r_a+c/r_c=b/r_b+d/r_d=cot(A'/2)+cot(B'/2)+cot(C'/2)+cot(D'/2) is obviousJankonyexnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-28094244666695378232010-05-08T00:29:17.539-07:002010-05-08T00:29:17.539-07:00E, F, G, H tg points of AB, BC, CD, AD
▲EOaA ~ ▲AH...E, F, G, H tg points of AB, BC, CD, AD<br />▲EOaA ~ ▲AHOd, ▲EOaB ~ ▲BFOb (OaOd bisector of kites)<br />=><br />AE/ra = AH/rd, EB/ra = BF/rb => <br />AE/ra + EB/ra = AH/rd + BF/rb<br /><br />a/ra = AH/rd + BF/rb (1)<br /><br />▲FObC ~ ▲CGOc, ▲COcD ~ ▲DHOd<br />FC/rb = CG/rc, GD/rc = DH/rd =><br />CG/rc + GD/rc = FC/rb + DH/rd<br /><br />c/rc = FC/rb + DH/rd (2)<br />from (1) and (2)<br />a/ra + c/rc = BF/rb + FC/rb + AH/rd + DH/rd<br /><br />A/ra + c/rc = b/rb + d/rd<br />--------------------------------------------c .t . e. onoreply@blogger.com