See complete Problem 111

Orthogonal Circles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Wednesday, May 28, 2008

### Geometry Problem 111

Labels:
chord,
circle,
circumcenter,
diameter,
orthogonal

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Angle COB = 180 - 2OCB, because ∆OCB is Isosceles. Angle BDC = 1\2COB = 90 - OCB.

ReplyDeleteAngle EFC = 2BDC. Angle ECF = (180 - EFC)\2 = OCB. Angle ACB = 90 because ∆ACB is a right trangle. therefore OCB + ACO =90, ACF = OCB therefore ACO + ACF = OCF = 90

Q.E.D

Is there a better way to prove this?

각BAC=각BDC=a

ReplyDelete각ACO=a

OC과 FC는 수직

각OCF=90도

Solution to Problem 111.

ReplyDeleteLet ang(BAC) = a.

We have ang(BDC) = a and ang(ACO) = a (because triangle AOC is isosceles). In the circle of center F, ang(CFE) = 2*ang(BDC) = 2a.

Then angle ACO is inscribed in arc EC, OC is tangent to the F circle and perpendicular to the radius CF. So ang(OCF) = 90º.

Join CB.

ReplyDeleteCF is tangent to circle (O) at C and CA is a chord of circle (O)

∴ ∠FCA = angle in the alternate segment = ∠CBA

∆OAC is isosceles.

∴ ∠OCA = ∠OAC = ∠BAC

Hence ∠OCF = ∠FCA + ∠OCA = ∠CBA + ∠BAC = 90° (∵∠ACB = 90°)

To Pravin: I have a doubt on your solution of problem 111. The fact that CF is tangent to circle (O) is not given in the enunciate. Besides, if CF is tangent to circle O, then it is perpendicular to the radius OC and ang(OCF) - 90º, this would be imediate.

ReplyDeleteTo Nilton Lapa: Thank you!. Please see the following Revised Proof.

ReplyDelete∠EDC = ∠BDC = ∠BAC (angles in the same segment)

∠BAC = ∠OAC = ∠OCA (since OA = OC)

∴ ∠EDC = ∠OCA

Next in the circle (F) through D, E, C

∠EDC = (1/2)∠EFC (angle subtended by arc EC at circumference = half the angle subtended by arc EC at the centre F)

Join EF. ∆ECF is isosceles (∵ FE = FC). So ∠CEF = ∠ECF

∠EFC = 180° - ∠CEF - ∠ECF = 180° - 2∠ECF

∴ ∠OCA = ∠EDC = (1/2)∠EFC = 90° - ∠ECF

Hence ∠OCF = ∠OCA + ∠ECF = 90°

If < EFC = 2@ then < EDC = @ = < BAC = OCA

ReplyDeleteSimilarly if < EFD = 2€ then < ACD = €

So < OCD = @+€

But in isoceles Tr. CDF, < DCF = 90 - @-€

So < OCF = 90

Sumith Peiris

Moratuwa

Sri Lanka