tag:blogger.com,1999:blog-6933544261975483399.post4933709574459139082..comments2024-03-29T17:38:24.729-07:00Comments on Go Geometry (Problem Solutions): Problem 362. Circle, Chord, Perpendicular, Equal chordsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-87249807440260044952017-10-18T13:36:31.237-07:002017-10-18T13:36:31.237-07:00Join AD and consider the isosceles triangle ACD. D...Join AD and consider the isosceles triangle ACD. Drop a perpendicular from C to AD and denote it as P<br />The points A,C,E and P are concyclic since m(AEC) = m(APC) = 90 degrees<br /><br />Join CB and let m(BAC)=X,m(BAD)=Y <br />=> m(DAC)=m(CDA)=X+Y<br />From angles in the same segment we have m(BDC)=X, m(DCB)=Y and m(CBA)=X+Y<br /><br />Since ACEP is concylic, => m(EPC)=m(EAC)=m(BAC)=X and m(PCE)=m(PAE)=m(DAE)=Y<br /><br />So the triangles APC and BEC are similar (AAA)<br />=> BE = AP.EC/PC ------------(1)<br /><br />Similarly, triangle PEC and DBC are similar<br />=> BD = EP.CD/PC = PE.AC/PC -----------(2) (since CD=AC)<br /><br />(1)+(2) => BE+BD = (AP.EC+PE.AC)/PC -----------(3)<br /><br />Applying ptolmey's to the cyclic quadrilateral ACEP<br />=> AE.PC = AP.EC+AC.PE <br />=> AE = (AP.EC+PE.AC)/PC -----------(4)<br /><br />From (3) and (4), BE+BD=AE Q.E.DSailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-52757395263386556892015-07-11T23:01:58.919-07:002015-07-11T23:01:58.919-07:00Extend AB to F such that AC = CF. Then C is the ci...Extend AB to F such that AC = CF. Then C is the circumcentre of Tr. ADF. Hence < DCF = 2.< DAF and since< DAC =< DCB it follows that BC bisects< DCF<br />Hence Tr.s CDB & CFB are congruent and so BD=BF<br />But AE=EF hence AE = EB + BD<br /><br />Sumith Peiris<br />Moratuwa<br />Sri Lanka Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-64650443036223704012009-12-20T08:34:55.112-08:002009-12-20T08:34:55.112-08:00join C with B, draw CFL ( F on AB, L on circle) th...join C with B, draw CFL ( F on AB, L on circle) that<br />FE = EB ( get ang CBE = ang CFB )<br /><br />now have to prove AF = DB<br /><br />ang CBE = ang ALF (at the same harc AC )<br />=> ang ALF = ang AFL ( AFL = CFE = CBE )<br />=> AL = AF<br /><br />now have to prove AL = BD<br /><br />ang FLD = CBE ( on equal harcs AC=CBD)<br />=>ang FLD = ang AFL<br />=> AB // LD, => AL = BD, or AF = BDc .t . e. onoreply@blogger.com