Saturday, June 1, 2013

Problem 882: Triangle, Angle, x, 2x, 7x, Cevian, Congruence, Auxiliary Lines

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 882.

Online Geometry Problem 882: Triangle, Angle x, 2x, 7x, Cevian, Congruence, Auxiliary Lines


  1. Ubicamos E en el segmento AC, tal que AD=DE=EC, entonces el ángulo EDB=4x. Luego trazamos el segmento EB y la bisectriz del ángulo ECB que es CF (F en EB), entonces este segmento es parte de la mediatriz del segmento EB. También proyectamos el punto E sobre el segmento AB (punto G) y trazamos el segmento EG, con esto logramos tres triángulos congruentes: DGE, CFE y CFB. Estos tres triángulos congruentes forman el triángulo notable BGE, donde el ángulo GBE mide 30 grados sexagesimales. Así, el ángulo BEC mide 30+2x grados; finalmente en el triángulo EFC tenemos que 30+6x=90 de donde x=10 grados.

  2. Draw an arc with radius BC from point C intersects AC and AB on point E and F respectively. Triangle BFC and GBC are congruent triangle then angle FBC = angle GCB = 8x.
    By the sum of interior angles theorem on triangle ABC:
    8x + 8x + 2x = 180
    x = 10.

    1. Where is point G located?

    2. the given property AD=BC is not used.
      by the way, where is point G in your solution?

  3. To Falensius

    I note that your solution never use the given information AD=BC .Is it true that AD=BC is not require to get x=10 ?
    Peter Tran

  4. On line AC the length is equal to BC.

  5. I've never used the information AD = BC. I just add more point and lines to solve the problem without considering the information. I will try to find another solution to solve the problem with considering the information AD = BC.

    thank u.

    1. To Falensius. In your solution of problem 882, why triangles BFC and GBC are congruent?

  6. Trigonometry solution:
    let put AC=1
    in triangle ACD, AD=sin(x)/sin(3x)
    in triangle ABC, BC=sin(2x)/sin(10x)
    equation AD=BC have 3 solutions, x=0, x=10 and x=45
    x=0 correspond to AD=BC=0
    x=10 correspond to AD=BC= finite number
    x=45 correspond to AD=BC= infinity

  7. See my solution, mentioned below:

  8. Let CM be the bisector of < ACB, M on AB.
    Let MN be drawn perpendicular to DC, N on AC

    Then Tr. s DMC, BMN, BCN, CND are all isoceles

    So Tr.s DMN, CMN and BCM are all congruent SSS.

    So < DMN = < CMN = < CMB = 6x so 6x + 6x+ 6x = 180 and x = 10

    Sumith Peiris
    Sri Lanka