tag:blogger.com,1999:blog-6933544261975483399.post4748042698521068371..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 116Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-45790923339386210142010-04-08T20:46:46.036-07:002010-04-08T20:46:46.036-07:00BE = (s-c)
[BPE]=1/2.rc.(s-c)=1/2.([ABC]/(s-c).(s-...BE = (s-c)<br />[BPE]=1/2.rc.(s-c)=1/2.([ABC]/(s-c).(s-c)<br /> =1/2.([ABC]<br /><br />BD = (s-a)<br />[BDQ]=1/2.ra.(s-a)=1/2.([ABC]/(s-a).(s-a)<br /> =1/2.([ABC]<br /><br />So [BPE] =[BDQ]<br /> [BPE]-[BGFH]=[BDQ]-[BGFH]<br /> S1+S2 = S3+S4<br /><br />widarto teddy<br />wteddyad@yahoo.comAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-26564312360381379922010-01-05T01:00:39.321-08:002010-01-05T01:00:39.321-08:00S1+S2+[BGFH]=[PBE]
S3+S4+[BGFH]=[DBQ]
[PBE]=0.5BP....S1+S2+[BGFH]=[PBE]<br />S3+S4+[BGFH]=[DBQ]<br />[PBE]=0.5BP.BE.sin(PBE)<br />[DBQ]=0.5DB.BQ.sin(DBQ)<br />line PBQ is the exterior bisector of ang(ABC)<br />ang(PBE)=(180-B)/2+ang(DBE)=ang(DBQ)<br />The right triangles PBD and BQE are similar<br />BP.BE=DB.BQ<br />hence [PBE]=[DBQ] and S1+S2=S3+S4<br />.-.Anonymousnoreply@blogger.com