Giải thích các bước giải:
Vì $\widehat{BOC}=120^o$
$\rightarrow\widehat{BAC}=60^o$
$\rightarrow\widehat{BAE}=\widehat{BAC}=\widehat{CAD}=60^o$
$\rightarrow A,D,E $ thẳng hàng
Lại có:
$\widehat{ABK}+\widehat{ADC}=180^o$
$\widehat{ADC}=\widehat{ABC}=\widehat{ABE}$
$\rightarrow\widehat{ABE}+\widehat{ABK}=180^o\rightarrow E,B,K$ thẳng hàng
Chứng minh tương tự ta có D,C,K thẳng hàng
Ta có:
$\widehat{BAK}=\widehat{BDK}=\widehat{DBC}=90^o-\widehat{ACB}=\widehat{BAO}$
$\rightarrow A,O,K$ thẳng hàng
Lại có:
$\widehat{EAB}=\widehat{EKD}\rightarrow\widehat{EKD}=60^o$
$\rightarrow cos\widehat{BKC}=\dfrac{1}{2}$
$\rightarrow cos\widehat{BKC}=\dfrac{KB^2+KC^2-BC^2}{2KB.KC}=\dfrac{1}{2}$
$\rightarrow KB^2+KC^2-BC^2=KB.KC$
$\rightarrow KB^2+KC^2=KB.KC+BC^2\ge 2KB.KC\rightarrow KB.KC\le BC^2$
$\rightarrow \dfrac{1}{KB^2}+\dfrac{1}{KC^2}\ge 2\sqrt{\dfrac{1}{KB^2}.\dfrac{1}{KC^2}}=\dfrac{2}{KB.KC}\ge \dfrac{2}{BC^2}$
Dấu = xảy ra
$\rightarrow KB=KC\rightarrow\Delta KBC$ đều
$\rightarrow KO\perp BC\rightarrow AO\perp BC$