## Saturday, November 14, 2009

### Problem 386. Triangle, Angle bisector, Perpendicular, Parallel, Semiperimeter, Congruence

Proposed Problem
Click the figure below to see the complete problem 386.

Complete Problem 386
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

1. by problem 388, GD || BE.
by problem 387, BEDG is cyclic.
So, BEDG is an isosceles trapezium.
SO,
BG = DE

2. BEFD is concyclic
BEDG 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

3. 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.

4. To Ramprakash.K
It is necessary E, D, M, to be collinear
Just draw the altitude of FGC from G (see arcs & EM = MC
Thanks

5. To Ramprakash.K
E, 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

6. 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???

7. To Ramprakash.K
EM = 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)

8. Kindly refer my proof to Problem 385

Sumith Peiris
Moratuwa
Sri Lanka