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.

 Problem 386. Triangle, Angle bisector, Perpendicular, Parallel, Semiperimeter.
See also:
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.
    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

  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
    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
    Sri Lanka