Geometry Problem

Click the figure below to see the complete problem 480 about Triangle, Circle, Center, Altitude, Angle, Circumcenter.

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Complete Problem 480

Level: High School, SAT Prep, College geometry

## Monday, July 26, 2010

### Problem 480: Triangle, Circle, Center, Altitude, Angle, Circumcenter

Labels:
altitude,
angle,
center,
circle,
circumcenter,
perpendicular,
triangle

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join A to O, draw OT perpendicular to AB ( T on AB )

ReplyDeleteang BCD = 90° - ß = 1/2 arc ACB (1)

ang AOB = arc ACB =>

ang AOT = 1/2 arc ACB (2)

from (1) & (2)

ang BCD = ang AOT =90°-α =90°-ß (3)( AT=TB,OT perpen )

from (3) =>

α = ß

I take the E point of maximum arc AB (not arc ACB).Is <BCD=<AEB=γ,then β+γ=90.

ReplyDeleteBut <AOB=2.<AEB=2γ,so 2α+2γ=180 or α+γ=90. Therefore α=β.

1. Draw in AO.

ReplyDelete2. Note triangles ABO ACO and BCO are all isosceles.

3. Let angle OBA = x and angle ABC = y (greek letters are hard in comments)

4. Then angle OAB = OBA = x from the isosceles triangle.

5. Angle OCB = OBC must be x + y from its isosceles triangle

6. Angle AOC is 2y since its the central angle and ABC is the inscribed one.

6. Angle OAC = ACO = 90 - y since this is also isosceles.

7. Angle BCD = 180 - (angle ACO + angle OCB) = 90 - x

8. Therefore CBD = 180 - 90 - BCD = x.

In general, ortocenter and circuncenter are isogonal conjugates.

ReplyDelete