Sunday, October 18, 2009

Problem 369. Intersecting circles, Chord, Center, Angle, Congruence.

Proposed Problem
Click the figure below to see the complete problem 369 about Intersecting circles, Chord, Center, Angle, Congruence.

Problem 369. Intersecting circles, Chord, Center, Angle, Congruence.
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Complete Geometry Problem 369
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 10:17 AM
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4 comments:

  1. Valdir Marques Faria, Colégio VisãoOctober 20, 2009 at 2:13 PM

    Como OC = AO, então o triângulo ACO é isósceles. Assim, Ang(OCA) = Ang(OAC). Os ângulo OBD e OAD são congruentes, pois seus vértices pertencem ao mesmo arco capaz. Como OC = OB e podemos ter que Ang(OCB) = Ang(OBC). Assim, Ang(BCO) + Ang(OCD) = Ang(CBO) + Ang(OBD), o que podemos concluir que o triângulo DCB é isósceles de base BC. Assim BD = CD (demonstrado).

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  2. Sumith PeirisJuly 8, 2015 at 4:44 AM

    < ADB = < AOB = 2 < ACB implying that CD= DB

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  3. Peter TranOctober 13, 2016 at 11:38 AM

    https://goo.gl/photos/V1K1LviHSrgYLvjy8

    Connect OA, OB, OC
    Observe that u=∠ODB=∠OAD=∠OCA
    And v=∠OBC=∠OCB
    And ∠ DBC=u+v = ∠DCB=> triangle CDB is isosceles => DB=DC

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