Proposed Problem

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

See more:

Complete Geometry Problem 369

Level: High School, SAT Prep, College geometry

## Sunday, October 18, 2009

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

Labels:
angle,
center,
chord,
congruence,
intersecting circles

Subscribe to:
Post Comments (Atom)

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).

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

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka

https://goo.gl/photos/V1K1LviHSrgYLvjy8

ReplyDeleteConnect OA, OB, OC

Observe that u=∠ODB=∠OAD=∠OCA

And v=∠OBC=∠OCB

And ∠ DBC=u+v = ∠DCB=> triangle CDB is isosceles => DB=DC