Giải thích các bước giải:

Gọi $A_1C_1\cap B_1D_1=E$

Ta có $A_1,B_1,C_1,D_1$ nằm giữa cung $AB,BC,CD,DA$

$\to A_1A=A_1B, B_1B=B_1C, C_1C=C_1D, D_1D=D_1A$

$\to\widehat{A_1EB_1}=\widehat{EA_1D_1}+\widehat{ED_1A_1}$

$\to\widehat{A_1EB_1}=\widehat{C_1A_1D_1}+\widehat{B_1D_1A_1}$

$\to\widehat{A_1EB_1}=(\widehat{C_1A_1D}+\widehat{DA_1D_1})+(\widehat{B_1D_1B}+\widehat{BD_1A_1})$

$\to\widehat{A_1EB_1}=(\dfrac12\widehat{C_1OD}+\dfrac12\widehat{DOD_1})+(\dfrac12\widehat{B_1OB}+\dfrac12\widehat{BOA_1})$

$\to\widehat{A_1EB_1}=\dfrac12(\widehat{C_1OD}+\widehat{DOD_1}+\widehat{B_1OB}+\widehat{BOA_1})$

$\to\widehat{A_1EB_1}=\dfrac12(\widehat{C_1OC}+\widehat{AOD_1}+\widehat{B_1OC}+\widehat{AOA_1})$

Tương tự chứng minh được:

$\widehat{A_1ED_1}=\dfrac12(\widehat{C_1OC}+\widehat{AOD_1}+\widehat{B_1OC}+\widehat{AOA_1})$

$\to \widehat{A_1EB_1}=\widehat{A_1ED_1}$

Mà $\widehat{A_1EB_1}+\widehat{A_1ED_1}=180^o$

$\to \widehat{A_1EB_1}=\widehat{A_1ED_1}=90^o$

$\to A_1C_1\perp B_1D_1$