tag:blogger.com,1999:blog-6933544261975483399.post7104750445077902542..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1164: Triangle, Angle Bisector, CevianAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-5317818881405378652017-11-18T21:08:43.471-08:002017-11-18T21:08:43.471-08:00Use Stewarts Law with triangle ABC and the cevian ...Use Stewarts Law with triangle ABC and the cevian BD:<br /><br />AB = 3x, AD = x, DE = 2x, CE = 6x<br /><br />(3x)^2*8x + BC^2*x = 9x(BD^2 + 2x^2)<br />That simplifies to BC^2 = 9BD^2 or BC = 3BD<br /><br />Since BC:BD = 3:1 and CE:DE = 3:1 also BE is the angle bisector by the angle bisector theorem.<br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />Benjamin Leishttps://www.blogger.com/profile/10974191081762367425noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-10177981317001065962016-01-31T15:07:26.305-08:002016-01-31T15:07:26.305-08:00See relations between tang and secant
Ang ABD = 1/...See relations between tang and secant<br />Ang ABD = 1/2 arc BD, Ang ACB = 1/2 arc BD<br />or Tr ABD similar to Tr ABCc.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57636144490193154032016-01-31T13:25:09.687-08:002016-01-31T13:25:09.687-08:00Yes, but why ¿ABD=ECB?, because:
The triangle ▲AB...Yes, but why ¿ABD=ECB?, because: <br />The triangle ▲ABD is proportional to the triangle ▲BCA<br />BAC = BAD (They have in common the angle A)<br />Its sides are proportional:AC/AB=AB/AD=3Ana de la Fuentehttps://www.blogger.com/profile/13473143575384692051noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-51707766199652535132015-11-13T18:07:42.466-08:002015-11-13T18:07:42.466-08:00Excellent simple solution! Excellent simple solution! Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-35785050448152607942015-11-13T10:11:14.971-08:002015-11-13T10:11:14.971-08:00Tr BAE isoceles => ang ABE=ang AEB => ABD ...Tr BAE isoceles => ang ABE=ang AEB => ABD + DBE = EBC + ECB <br /> => DBE = EBC ( ABD = ECB )c.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-24368288710415487702015-11-12T23:26:43.211-08:002015-11-12T23:26:43.211-08:00Let AD = p so that DE = 2p, CE = 6p and AB = 3p
N...Let AD = p so that DE = 2p, CE = 6p and AB = 3p<br /><br />Now AB^2 = 9p^2 = AD.AC so AB is tangential to DBC at B<br /><br />From similar Tr.s AD/AB = BD/BC = 1/3 = 2p/6p = DE/EC <br /><br />Hence BE bisects < DBC<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-45019484938181042492015-11-12T16:35:20.709-08:002015-11-12T16:35:20.709-08:00Let AD=x so DE=2x, EC=6x and AB=3x
calculate BD an...Let AD=x so DE=2x, EC=6x and AB=3x<br />calculate BD and BC using cosine formula in triangles ABD and ABC .we get<br />BD^2=10x^2-6x^2.cos(A)<br />BC^2=90x^2-54x^2.cos(A)<br />(BD/BC)^2= 1/9 => BD/BC=1/3= ED/EC<br />so BE bisect angle DBCPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com