Cho hình chóp $S.ABC$ có đáy $ABC$ là tam giác đều cạnh $a$...

Xét \)">\Delta SAB \Delta SCB\widehat{SAB}=\widehat{SCB}=90{}^\circ ; AB=BCSB\Delta SAB=\Delta SCBSABAE\bot SBCE\bot SB\left( \widehat{\left( SAB \right) , \left( SBC \right)} \right)=\left( \widehat{AE,CE} \right)=60{}^\circ \widehat{AEC}=\left( \widehat{AE,CE} \right)=60{}^\circ AE=AC=AB=aAEBE\widehat{AEC}=180{}^\circ -\left( \widehat{AE,CE} \right)=120{}^\circ AECEEK\widehat{EAK}=30{}^\circ AEKAE=\dfrac{AK}{\cos 30{}^\circ }=\dfrac{\sqrt{3}}{3}aABEBE=\sqrt{A{{B}^{2}}-A{{E}^{2}}}=\sqrt{{{a}^{2}}-\dfrac{{{a}^{2}}}{3}}=\dfrac{\sqrt{6}}{3}aSABBS=\dfrac{A{{B}^{2}}}{BE}=\dfrac{a\sqrt{6}}{2}{{V}_{B.EAC}}=\dfrac{1}{3}.BE.\dfrac{1}{2}.EA.EC.\sin 120{}^\circ =\dfrac{1}{3}.\dfrac{a\sqrt{6}}{3}.\dfrac{1}{2}.{{\left( \dfrac{a}{\sqrt{3}} \right)}^{2}}.\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{2}{{a}^{3}}}{36}\dfrac{{{V}_{B.EAC}}}{{{V}_{B.SAC}}}=\dfrac{BE}{BS}.\dfrac{BA}{BA}.\dfrac{BC}{BC}=\dfrac{BE}{BS}=\dfrac{\dfrac{a\sqrt{6}}{3}}{\dfrac{a\sqrt{6}}{2}}=\dfrac{2}{3}\Rightarrow {{V}_{B.SAC}}=\dfrac{3}{2}.{{V}_{B.EAC}}=\dfrac{3}{2}.\dfrac{\sqrt{2}}{36}{{a}^{3}}=\dfrac{\sqrt{2}}{24}{{a}^{3}}{{V}_{S.ABC}}=\dfrac{\sqrt{2}}{24}{{a}^{3}}$.