tag:blogger.com,1999:blog-6933544261975483399.post1304812719514565770..comments2022-09-27T03:11:10.165-07:00Comments on Go Geometry (Problem Solutions): Archimedes' Book of Lemmas, Proposition #13Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-33500865046681808382016-10-23T00:07:06.542-07:002016-10-23T00:07:06.542-07:00Let AF meet the circle at H.
DHBC is an isoceles ...Let AF meet the circle at H.<br /><br />DHBC is an isoceles trapezoid and HFGB is a rectangle. <br /><br />Triangles BGC & HFD are therefore congruent ASA.<br /><br />Hence DF = CG.<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-25803993736520635112013-06-25T16:13:10.467-07:002013-06-25T16:13:10.467-07:00http://img43.imageshack.us/img43/8577/u2mk.pnghttp://img43.imageshack.us/img43/8577/u2mk.pngAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-34939059462843616542011-09-08T07:36:13.597-07:002011-09-08T07:36:13.597-07:00produce BG to cut the circle at P,=>angle BAP=9...produce BG to cut the circle at P,=>angle BAP=90<br />If we observe AFPG, angles P,G,F are 90 degrees<br />=>angle FAP is 90 degrees,therefore AFPG is rectangle =>AF=PG<br />Now triangles AFD and PGC are similair and AF=PG<br />=>triangles AFD and PGC are congruent<br />=>CG=FDPradyumnanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-35249469849645545632011-08-07T04:27:58.033-07:002011-08-07T04:27:58.033-07:00Draw OH ⊥ CD. OH bisects CD
CH = HD
AF ∥ OH ∥ BG
T...Draw OH ⊥ CD. OH bisects CD<br />CH = HD<br />AF ∥ OH ∥ BG<br />Transversal AB is bisected at O<br />So transversal GF is bisected at H<br />GH = HF<br />Follows<br />CG = CH - GH = HD - HF = FDPravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-73590975234511847032011-08-04T22:16:11.433-07:002011-08-04T22:16:11.433-07:00AD = ABcos(BAD) & DF=ADcos(ADC) or DF=ABcos(AD...AD = ABcos(BAD) & DF=ADcos(ADC) or DF=ABcos(ADC)cos(BAD) But /_ADC=/_ABC (angles in the same sector) & /_BCD=/_BAD for the same reason. Hence DF=ABcos(ABC)cos(BAD)=BCcos(BCD)=CG<br />AjitAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-59454816092952224092011-07-31T23:34:18.986-07:002011-07-31T23:34:18.986-07:00Let O is the center of circle and I is the project...Let O is the center of circle and I is the projection of O over CD.<br />Since O is the midpoint of AB so I is the midpoint of FG and DC( F, I, G are the projection of A,O,B)<br />CG=CI-IG<br />DF=ID-IF <br />Since ID=IC and IF=IG so CG=DF<br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-46736315855601345892009-06-11T09:26:36.392-07:002009-06-11T09:26:36.392-07:00The situation becomes clear, if you extend the lin...The situation becomes clear, if you extend the lines AF and BG, so that they intersect the circle a second time.<br /><br />Let then AF intersect the circle in P and let BG intersect the circle in Q. Then we have angles BPA=AQB=90° (Thales). As AP and BQ are parallel, QBA=PAB and ABP=BAQ as they are alternate angles; and as the sum of them is 180° (they are part of a cyclic quadrangle), each of them is 90°.<br /><br />Thus APBQ is a rectangle. If we now draw the line through the center of the circle, which is perpendicular to CD, this will be a median of the rectangle (as a parallel to two sides which passes through the center). So, if we consider a reflection in this line, the circle and the rectangle are invariant and the segments DF and CG are mapped to each other. Thus they are equal. (by Thomas)Anonymousnoreply@blogger.com