Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1179.
http://s9.postimg.org/71ehki32n/pro_1179.pngDraw median BM and lines AL//DE and CN // DENote that triangles ALM and CNM are congruence… ( case ASA)We have a1/a2= GL/GB=(GM-ML)/GB= ½ - ML/ GB …. (1)C1/c2= GN/GB= ( GM+MN)/GB = ½ + MN/GB …. (2)Add ( 1) to (2) and note that ML=MN we get the result.
MANOLOUDIS APOSTOLIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECEBG intersecting AC M is AM=MC and BG=2.MG ,si DE intersecting AC K .Apply the theorem Menelaus tri ABM and tri BMC intersecting by KDGE.Therefore AK/KM.MG/GB.BD/DA=1 or DA/DB=AK/2.KM and MK/KC.EC/EB.BG/GM=1 orEC/EB=KC/2.MK. Therefore DA/DB+EC/EB=AK/2.MK+KC/2.MK=(AK+KC)/2.MK=(MK-AM+KM+MC)/2.MK=2.MK/2.MK=1.