tag:blogger.com,1999:blog-6933544261975483399.post1215609579447658826..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1512: Finding the Length of a Segment in a Triangle with a Median and a Cevian with Given Ratio. Difficulty Level: High School.Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-89580793379875727902023-10-12T12:36:10.694-07:002023-10-12T12:36:10.694-07:00From Rimitti / Wahran / Algeria
Apply Menelau'...From Rimitti / Wahran / Algeria<br />Apply Menelau's theorem to triangle EBC and secant AFD :<br />(BD/DC)*(CA/AE)*(EF/FB) = 1 with CA= 3AE and BD=DC ---> 3*EF/FB = 1---> FB =18Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-11205391245310873002023-02-25T23:19:43.506-08:002023-02-25T23:19:43.506-08:00If G is the mid point of BE, from the mid point th...If G is the mid point of BE, from the mid point theorem, <br />GD = CE/2 = AE<br />Then Tr.s AEF & DFG are congruent ASA <br /><br />So GF = EF = 6 and so GE = 6 + 6 = 12<br />Hence BF = BG + GF = GE + GF = 12 + 6 = 18<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-70637359529023738442023-02-25T01:12:38.326-08:002023-02-25T01:12:38.326-08:00https://photos.app.goo.gl/cDwW2cJUjC13tptQ9https://photos.app.goo.gl/cDwW2cJUjC13tptQ9c.t.e.o.noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-1577852297908572722023-02-24T18:21:31.492-08:002023-02-24T18:21:31.492-08:00https://photos.app.goo.gl/Rr1mPzhRSN29rhfb6
Let C...https://photos.app.goo.gl/Rr1mPzhRSN29rhfb6<br /><br />Let CF meet AB at G<br />Apply Ceva theorem for concurrent point F we have (EA/EC) x(DC/DB) x( GB/GA)= 1<br />Replace DC/DB= 1 and EA/EC=½ we get GB/GA= 2<br /><br />Apply Van Aubel theorem we have FB/FE= GB/GA + DB/DC = 2+1= 3<br />So BF= 3GA= 18<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com