Wednesday, November 30, 2011

Problem 695: Triangle, Angles, Auxiliary Construction, Congruence, Mind Map, Polya

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 695.

Online Problem 694: Triangle, Angles, 30 Degrees, Auxiliary Construction, Mind Map, Polya.

5 comments:

  1. extend BA to K
    BC bisector
    BD=BK
    Join C to K
    CK=CD
    see the KCDA cydric quatrilateral
    x=25

    ReplyDelete
  2. Problem 695 again:
    Let the circle through B, C, D cut AC at E.
    ∠EBD = ∠ECD = 27° and so ∠EBC = 90°.
    Let O be the midpoint of EC.
    OB = OE = OC and O is the centre of circle BCD
    ∠AOD = ∠EOD = 2∠ECD = 54° = ∠ABD
    So A,B,O,D are concyclic.
    Hence
    x = ∠DAO = ∠DBO
    = 63° - ∠OBC
    = 63° - ∠OCB
    = 63° - 38°
    = 25°

    ReplyDelete
  3. Let < ABD be bisected by BEF, E on AC and F on CD. Then BCDE and ABCF are both cyclic so since < EBC = 90, < EDC = 90 = < CAF which shows that AEDF is cyclic.

    Hence x = < EFD = 25

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Further we see that E is the incentre of Tr. ABD.

    The 3 sides of Tr. ECF are the diameters of the 3 cyclic quads in this problem

    ReplyDelete
  5. Problem 695
    Forms the isosceles triangle AΒΕ with ΒE=ΑΕ and <ABE=<BAE=52.( Point E located under the D ). Then <BEA=76=2.38=2.<BCA. So the point E is circumcenter triangle ABC. But
    <BAC=180-54-63-38=25, so <ECA=<EAC=52-25=27, (AE=BE=CE) is <DCA=27 then C,D,E
    Is collinear with <EBD=54-52=2 and <BDC=180-63-38-27=52=<BAE. So the ABDE is
    Cyclic .Then <EAD=<EBD=2.Therefore x=<DAC=27-2=25.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete