Proposed Problem
Click the figure below to see the complete problem 386.
See also:
Complete Problem 386
Collection of Geometry Problems
Level: High School, SAT Prep, College geometry
Saturday, November 14, 2009
Problem 386. Triangle, Angle bisector, Perpendicular, Parallel, Semiperimeter, Congruence
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by problem 388, GD || BE.
ReplyDeleteby problem 387, BEDG is cyclic.
So, BEDG is an isosceles trapezium.
SO,
BG = DE
BEFD is concyclic
ReplyDeleteBEDG is concyclic
=> BEFDG is concyclic (three points determine the circle
Draw EM median of BEC => ang MEC = ang ECM
=> ang MEC = ang ACE
=> ED//AC
if 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,
ReplyDeleteHow 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.K
ReplyDeleteIt is necessary E, D, M, to be collinear
Just draw the altitude of FGC from G (see arcs & EM = MC
Thanks
To Ramprakash.K
ReplyDeleteE, midpoint of BK, K on AC, D midpoint of BL, L on AC
=> ED//AC
D, M midpoints => DM//AC
From D two // to AC or E, D, M collinear
to c.t.e.o...
ReplyDeleteEM=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.K
ReplyDeleteEM = MC= MB = R ( radius of BEC )
=> MEC = ECM => MEC = ECA as alternate angles
EM//AC
------------------------------------
Or ED midle line of BKL
DM midle line of BLC
=> EM midle line of BKC
about E, D, M see comment 5 (I think is clear one)
Kindly refer my proof to Problem 385
ReplyDeleteSumith Peiris
Moratuwa
Sri Lanka