Problem: Triangle ABCA B CABC is inscribed in a circle with center Oβ²O'Oβ². A circle with center OOO is inscribed in triangle ABCA B CABC. AOA OAO is drawn, and extended to intersect the larger circle in DDD. Then we must have:
Answer Choices:
A. CD=BD=Oβ²DCD=BD=O' DCD=BD=Oβ²D
B. AO=CO=ODAO=CO=ODAO=CO=OD
C. CD=CO=BDCD=CO=BDCD=CO=BD
D. CD=OD=BDCD=OD=BDCD=OD=BD
E. Oβ²B=Oβ²C=ODO' B=O' C=ODOβ²B=Oβ²C=OD Solution:
Quadrilateral ABDC is inscriptible
β΄CD^=BD^,β΄CD=BD\therefore \widehat{C D}=\widehat{B D}, \quad \therefore \mathrm{CD}=\mathrm{BD}β΄CD=BD,β΄CD=BD \quad
β DCO=Ξ±+Ξ²,β DOC=Ξ±+Ξ²,β΄OD=CD=BD\angle \mathrm{DCO}=\alpha+\beta, \quad \angle \mathrm{DOC}=\alpha+\beta, \therefore \mathrm{OD}=\mathrm{CD}=\mathrm{BD}β DCO=Ξ±+Ξ²,β DOC=Ξ±+Ξ²,β΄OD=CD=BD \quad
