Kẻ AH ⊥ mp(P) và AI ⊥ BC
Thì \(\beta = \widehat {ABH},\gamma = \widehat {ACH},\alpha = \widehat {AIH}.\)
Vì ΔABC vuông ở A nên :
\(\eqalign{ & {1 \over {A{I^2}}} = {1 \over {A{B^2}}} + {1 \over {A{C^2}}} \cr & \Rightarrow {{A{H^2}} \over {A{I^2}}} = {{A{H^2}} \over {A{B^2}}} + {{A{H^2}} \over {A{C^2}}} \cr & hay\,\,{\sin ^2}\alpha = {\sin ^2}\beta + {\sin ^2}\gamma \cr} \)
