1 / 9

Central Angle : an angle with its vertex at the center of the circle

Inscribed and Central Angle. Central Angle : an angle with its vertex at the center of the circle. Inscribed Angle : an angle whose vertex on a circle and whose side contain chord of the circle. Created by ﺠﻴﻄ for Mathlabsky.wordpress.com.

myrna
Télécharger la présentation

Central Angle : an angle with its vertex at the center of the circle

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Inscribed and Central Angle Central Angle : an angle with its vertex at the center of the circle Inscribed Angle : an angle whose vertex on a circle and whose side contain chord of the circle Created by ﺠﻴﻄ for Mathlabsky.wordpress.com Created by ﺠﻴﻄ for Mathlabsky.wordpress.com

  2. Inscribed and Central Angle A B AOB = Central Angle ACB = Inscribed Angle O • b a C c Theorem a +  b =  c x Proof :

  3. AOB = 2 ACB A Let ACO = p, and BCO = q p O 2p p C D OC = OA (radius) ACO =  CAO q 2q q OC = OB (radius) BCO =  CBO B • b AOD = 2p & BOD = 2q a c AOB = 2p + 2q & ACB = p + q  AOB = 2 ACB

  4. ACB = ADB A B O • AOB = 2 ACB AOB = 2 ADB ACB = ADB C D “If two inscribed angles intercept the same arc, than the angle are congruent”

  5. ABC = 900 A AOC = 1800 B O AOC = 2 ABC • C ABC = 900 “An angle inscribed in a semicercle is a right angle”

  6. BAD + BCD = 180 A D 2 Let m1=2x 3 O  m2 = x and m3 = (360 – 2x) 1 4 C • m3 = (360 – 2x) B  m4 = ½(360 – 2x)  m4 = 180 – x m2 + m4 = x + (180 – x) BAD + BCD = 180 “If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary”

  7. A D 3 3 E 1 1 AEB = CED 2 2 ABD = ACD B • C BAC = BDC E E  ABE  CDE  1 1 3 2 2 3 B C A D

  8. AEB = ½ AOB+ ½ COD A Let AOB = x and COD = y D E  ACB = ½ x ● O y X C  DBC = ½ y • B A E ½ x C  AEB = ½ x + ½ y ½ y B AEB = ½ AOB+ ½ COD

  9. AEB = ½ AOB - ½ COD A Let AOB = x and COD = y D E O y x  ADB = ½ x C  DBC = ½ y • B A D ½ x E  AEB = ½ x - ½ y ½ y B AEB = ½ AOB - ½ COD

More Related