Online Geometry theorems, problems, solutions, and related topics.
See complete Problem 111Orthogonal Circles. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
Angle COB = 180 - 2OCB, because ∆OCB is Isosceles. Angle BDC = 1\2COB = 90 - OCB.Angle 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 = 90Q.E.DIs there a better way to prove this?
각BAC=각BDC=a각ACO=aOC과 FC는 수직각OCF=90도
Solution to Problem 111.Let 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.CF 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 = ∠BACHence ∠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.
To Nilton Lapa: Thank you!. Please see the following Revised Proof.∠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° - ∠ECFHence ∠OCF = ∠OCA + ∠ECF = 90°
If < EFC = 2@ then < EDC = @ = < BAC = OCA Similarly if < EFD = 2€ then < ACD = €So < OCD = @+€ But in isoceles Tr. CDF, < DCF = 90 - @-€So < OCF = 90Sumith PeirisMoratuwaSri Lanka