tag:blogger.com,1999:blog-6933544261975483399.post2405074326975345338..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1482: Right Triangle, Perpendicular, Double Angle, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-82674930657645989432022-01-07T08:42:15.028-08:002022-01-07T08:42:15.028-08:00Simple Solution
Extend EB to F such that EB = BF ...Simple Solution<br /><br />Extend EB to F such that EB = BF = 6 and AF = FE<br /><br />Tr.s ABE & ABC are similar and AE bisects < CAF and so <br />AE/DE = AC / CE = AF / FE = AE / 2.BE<br /><br />Hence DE = 2.BE = 12<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-80384626850950563902020-11-23T12:02:47.224-08:002020-11-23T12:02:47.224-08:00https://photos.app.goo.gl/XN6AessMnkA9GsNf6https://photos.app.goo.gl/XN6AessMnkA9GsNf6c.t.e.o.https://www.blogger.com/profile/16937400830387715195noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-17554310483016679302020-11-20T16:38:43.200-08:002020-11-20T16:38:43.200-08:00Fold Tr.ABE along AE such that B' is the refle...Fold Tr.ABE along AE such that B' is the reflection of B (ABEB' is cyclic )<br />m(BAE)=m(EAB')=Alpha=>AB' is angle bisector of m(EAD). Extend EB' to meet AC at F <br />Since AB' bisects m(EAF) and AB' perpendicular to EF => EF=2.EB'=2.BE=12<br />m(EFA)=180-m(EFC)=2Alpha+Theta=m(EDF)<br />=> EDF is an isosceles triangle and ED=EF=12<br />Sailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-33217686614253029302020-11-15T05:03:58.031-08:002020-11-15T05:03:58.031-08:00Extend EB to F such that BE = BF = 6 so that BA bi...Extend EB to F such that BE = BF = 6 so that BA bisects < FAE<br /><br />Then < FEA = 2.Alpha + Theta = 90 - Alpha from which easily < CED = 4.Alpha = < CAF<br /><br />Hence ADEF is concyclic and since EA bisects < CAF, it follows that DE = EF = 12<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76198445766166605272020-11-12T03:04:06.632-08:002020-11-12T03:04:06.632-08:00Let:
AB = a
1.)
3α + Θ +90 = 180 => Θ = 90 - 3...Let:<br />AB = a<br /><br />1.)<br />3α + Θ +90 = 180 => Θ = 90 - 3α<br /><br />2.)<br />Angle x / (2sin(α)) = √(a² + 36)<br />_________________<br /><br />From (3) and (4):<br /><br />x = 12.<br /><br />Ludwig Mercknoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76718858386071985652020-11-10T22:24:42.575-08:002020-11-10T22:24:42.575-08:00https://photos.app.goo.gl/wrmr2bHVba98ZA3k8
Draw ...https://photos.app.goo.gl/wrmr2bHVba98ZA3k8<br /><br />Draw angle bisector AH of angle CAE<br />Draw EF ⊥AH ( see sketch)<br />Note that ∠ (CEF)= 2. Alpha and triangle EAF is isosceles<br />Triangle ABE congruent to AEH and AFH<br />So EH=HF=BE=6 so EF=12<br />In triangle EFC , external angle ∠ (DFE)= theta+2. Alpha<br />In triangle AED, external angle ∠ (EDF)= theta+2. Alpha<br />So tringle DEF is isosceles and DE=EF=12<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-77258583892341661892020-11-09T02:26:20.018-08:002020-11-09T02:26:20.018-08:00Draw a)EH perpend AC b)AP bisector of angEAC c)...Draw a)EH perpend AC b)AP bisector of angEAC c)EMT perpend AP<br />=> Tr DET isosceles, from EM=6 => ET=DE=12c.t.e.o.https://www.blogger.com/profile/16937400830387715195noreply@blogger.com