Ta có: ΔOHP=ΔOHM (2 cạnh góc vuông bằng nhau)
⇒$\widehat{POH}=\widehat{MOH}$
Tượng tự: ΔMOK=ΔQOK (2 cạnh góc vuông bằng nhau)
⇒$\widehat{MOK}=\widehat{QOK}$
Ta có: $\widehat{POQ}=\widehat{POH}+\widehat{QOK}+\widehat{MOH}+\widehat{MOK}$
⇒$\widehat{POQ}=2(\widehat{MOH}+\widehat{MOK})$
⇔$\widehat{POQ}=2\widehat{KOH}=120^o$