tag:blogger.com,1999:blog-6933544261975483399.post121719565002010732..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 882: Triangle, Angle, x, 2x, 7x, Cevian, Congruence, Auxiliary LinesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-6933544261975483399.post-73308075740166906612015-11-10T09:59:42.489-08:002015-11-10T09:59:42.489-08:00Let CM be the bisector of < ACB, M on AB.
Let M...Let CM be the bisector of < ACB, M on AB.<br />Let MN be drawn perpendicular to DC, N on AC<br /><br />Then Tr. s DMC, BMN, BCN, CND are all isoceles <br /><br />So Tr.s DMN, CMN and BCM are all congruent SSS.<br /><br />So < DMN = < CMN = < CMB = 6x so 6x + 6x+ 6x = 180 and x = 10<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-77781109486564223632013-06-07T04:26:38.296-07:002013-06-07T04:26:38.296-07:00To Falensius. In your solution of problem 882, why...To Falensius. In your solution of problem 882, why triangles BFC and GBC are congruent?Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-61005599786201870022013-06-06T19:45:44.050-07:002013-06-06T19:45:44.050-07:00See my solution, mentioned below:
http://imageshac...See my solution, mentioned below:<br />http://imageshack.us/f/708/p882.png/Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-45058514043530731102013-06-06T16:56:33.650-07:002013-06-06T16:56:33.650-07:00Trigonometry solution:
let put AC=1
in triangle AC...Trigonometry solution:<br />let put AC=1<br />in triangle ACD, AD=sin(x)/sin(3x)<br />in triangle ABC, BC=sin(2x)/sin(10x)<br />equation AD=BC have 3 solutions, x=0, x=10 and x=45<br />x=0 correspond to AD=BC=0<br />x=10 correspond to AD=BC= finite number<br />x=45 correspond to AD=BC= infinity<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-28526342602757128222013-06-04T18:58:55.730-07:002013-06-04T18:58:55.730-07:00I've never used the information AD = BC. I jus...I've never used the information AD = BC. I just add more point and lines to solve the problem without considering the information. I will try to find another solution to solve the problem with considering the information AD = BC.<br /><br />thank u.Falensius Nangohttps://www.blogger.com/profile/15478259416020482717noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-33096390238684319732013-06-04T00:37:26.316-07:002013-06-04T00:37:26.316-07:00On line AC the length is equal to BC. On line AC the length is equal to BC. Falensius Nangohttps://www.blogger.com/profile/15478259416020482717noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-84857759385094088522013-06-03T21:20:40.596-07:002013-06-03T21:20:40.596-07:00To Falensius
I note that your solution never use ...To Falensius<br /><br />I note that your solution never use the given information AD=BC .Is it true that AD=BC is not require to get x=10 ?<br />Peter Tran<br /><br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-20379477259849169232013-06-03T21:12:37.738-07:002013-06-03T21:12:37.738-07:00the given property AD=BC is not used.
by the way, ...the given property AD=BC is not used.<br />by the way, where is point G in your solution?nakahttps://www.blogger.com/profile/11277356476170372732noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-64684038634813478112013-06-03T20:16:45.160-07:002013-06-03T20:16:45.160-07:00Where is point G located?Where is point G located?Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-67054416825692598722013-06-03T04:25:52.482-07:002013-06-03T04:25:52.482-07:00English please..English please..kalpithttps://www.blogger.com/profile/08431159710282150473noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-7800025976228792612013-06-03T01:58:24.802-07:002013-06-03T01:58:24.802-07:00Draw an arc with radius BC from point C intersects...Draw an arc with radius BC from point C intersects AC and AB on point E and F respectively. Triangle BFC and GBC are congruent triangle then angle FBC = angle GCB = 8x.<br />By the sum of interior angles theorem on triangle ABC:<br />8x + 8x + 2x = 180<br />x = 10.Falensius Nangohttps://www.blogger.com/profile/15478259416020482717noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-50057261890475051782013-06-02T11:08:59.040-07:002013-06-02T11:08:59.040-07:00Ubicamos E en el segmento AC, tal que AD=DE=EC, en...Ubicamos E en el segmento AC, tal que AD=DE=EC, entonces el ángulo EDB=4x. Luego trazamos el segmento EB y la bisectriz del ángulo ECB que es CF (F en EB), entonces este segmento es parte de la mediatriz del segmento EB. También proyectamos el punto E sobre el segmento AB (punto G) y trazamos el segmento EG, con esto logramos tres triángulos congruentes: DGE, CFE y CFB. Estos tres triángulos congruentes forman el triángulo notable BGE, donde el ángulo GBE mide 30 grados sexagesimales. Así, el ángulo BEC mide 30+2x grados; finalmente en el triángulo EFC tenemos que 30+6x=90 de donde x=10 grados.Anonymoushttps://www.blogger.com/profile/09887116460653340969noreply@blogger.com