Online Geometry theorems, problems, solutions, and related topics.
Proposed ProblemClick the figure below to see the complete problem 386.See also:Complete Problem 386Collection of Geometry ProblemsLevel: High School, SAT Prep, College geometry
by problem 388, GD || BE.by problem 387, BEDG is cyclic.So, BEDG is an isosceles trapezium.SO,BG = DE
BEFD is concyclicBEDG is concyclic=> BEFDG is concyclic (three points determine the circleDraw EM median of BEC => ang MEC = ang ECM=> ang MEC = ang ACE => ED//ACif extend BE to K on AC => ang BKC = ang DEB = ang EBC (ED//AC & BKC isosceles=> BEDG trapezoid concyclic=> BG = ED
to c.t.e.o,How do you know that D lies on the median from E to BC (in triangle BEC).I guess u can only conclude that EM||AC.
To Ramprakash.KIt is necessary E, D, M, to be collinearJust draw the altitude of FGC from G (see arcs & EM = MCThanks
To Ramprakash.KE, midpoint of BK, K on AC, D midpoint of BL, L on AC=> ED//ACD, M midpoints => DM//ACFrom D two // to AC or E, D, M collinear
to c.t.e.o...EM=MC. =>EM || AC.How exactly do you prove that E,D,M are collinear???could you give me a full proof of it, by your method, please???
To Ramprakash.KEM = MC= MB = R ( radius of BEC )=> MEC = ECM => MEC = ECA as alternate anglesEM//AC------------------------------------Or ED midle line of BKLDM midle line of BLC=> EM midle line of BKCabout E, D, M see comment 5 (I think is clear one)
Kindly refer my proof to Problem 385Sumith PeirisMoratuwaSri Lanka