tag:blogger.com,1999:blog-6933544261975483399.post729473818924147519..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 546: Triangle, Angle Trisectors, 60 Degrees, Equilateral triangleAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-33580306204540369562015-10-01T08:33:17.059-07:002015-10-01T08:33:17.059-07:00Let CE meet AB at M and let AD and CF meet at N
N...Let CE meet AB at M and let AD and CF meet at N<br /><br />Now < AEM = a+c so < MED = 60 - a, hence < DEF must be 60 (< CEF = 60 + a)<br /><br />Further < ADE = < EFC = 120 - a - c so < EDN = < EFN <br /><br />Also E is the incentre of Tr. ANC hence < DNE = < FNE<br /><br />So Tr. s DEN and FEN are congruent ASA<br /><br />So in Tr. DEF. DE = EF and the included angle is 60 hence the Tr. DEF is equilateral<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-90821795763862256492010-12-04T12:12:26.387-08:002010-12-04T12:12:26.387-08:00DEF = 60°, ADE = CFE = 120° - a - c
AD meet CF at...DEF = 60°, ADE = CFE = 120° - a - c<br />AD meet CF at H => HE bisector<br />build EAG = c, G on CF, join E to G <br />▲ADE ≡ ▲EFG (DAE = FGE, AE = EG, AED = GEF )<br />=> ED = EF => EDF = EFD = 60°c .t . e. onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-28120191608991146312010-12-03T11:04:52.761-08:002010-12-03T11:04:52.761-08:001)in triangle AEC
<AEC=180-a-c=y
the sine law g...1)in triangle AEC<br /><AEC=180-a-c=y<br />the sine law gives:EC=ACsina/siny;EA=ACsinc/siny<br />2)at vertex E<br /><DEF=360-y-60-a-60-c=60<br />3)in triangle ADE<br /><ADE=180-a-c=x<br />EDsinx=AEsina<br />4)in triangle CEF<br /><CEF=x<br />ECsinc=EFsinx<br />5)DE=EF=ACsina.sinc/(siny.sinx)<br /><DEF=60,DEF is equilateral<br />.-.Anonymousnoreply@blogger.com