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

Labels:
angle bisector,
congruence,
parallel,
perpendicular,
triangle

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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