Cho hình lăng trụ đứng $ABC.{A}'{B}'{C}'$ có đáy là tam giác $ABC$...

Gọi là trung điểm của ta có

Trong kẻ

Ta có:
$\left\{ \begin{array}{*{35}{r}}
\left( A{B}'C \right)\cap \left( BC{C}'{B}' \right)={B}'C \\
\left( A{B}'C \right)\supset AD\bot {B}'C \\
\left( BC{C}'{B}' \right)\supset HD\bot {B}'C \\
\end{array} \right.$$\left. \Rightarrow \left( \left( A{B}'C \right)\widehat{;(B}C{C}'{B}' \right) \right)=\left( A\widehat{D;H}D \right)=\widehat{ADH}ABCAHAH=\dfrac{BC}{2}=\dfrac{a\sqrt{2}}{2}AHDHD=AH.\text{cot}60{}^\circ =\dfrac{a\sqrt{6}}{6}\Delta CB{B}'\!\!\Delta\!\!CDH\Rightarrow \dfrac{B{B}'}{HD}=\dfrac{C{B}'}{CH}\Rightarrow \dfrac{B{B}'}{\dfrac{a\sqrt{6}}{6}}=\dfrac{\sqrt{2{{a}^{2}}+B{{{{B}'}}^{2}}}}{\dfrac{a\sqrt{2}}{2}}\Leftrightarrow \sqrt{3}B{B}'=\sqrt{2{{a}^{2}}+B{{{{B}'}}^{2}}}\Leftrightarrow 2B{{{B}'}^{2}}=2{{a}^{2}}\Leftrightarrow B{B}'=aABCA{{B}^{2}}+A{{C}^{2}}=B{{C}^{2}}\Leftrightarrow AB=AC=\dfrac{BC}{\sqrt{2}}=a\Rightarrow {{S}_{ABC}}=\dfrac{1}{2}AB.AC=\dfrac{{{a}^{2}}}{2}\Rightarrow {{V}_{ABC.{A}'{B}'{C}'}}=B{B}'.{{S}_{ABC}}=a.\dfrac{{{a}^{2}}}{2}=\dfrac{{{a}^{3}}}{2}{{V}_{A{B}'C{A}'{C}'}}+{{V}_{{B}'.ABC}}={{V}_{ABC.{A}'{B}'{C}'}}\Rightarrow {{V}_{A{B}'C{A}'{C}'}}={{V}_{ABC.{A}'{B}'{C}'}}-{{V}_{{B}'.ABC}}={{V}_{ABC.{A}'{B}'{C}'}}-\dfrac{1}{3}{{V}_{ABC.{A}'{B}'{C}'}}=\dfrac{2}{3}{{V}_{ABC.{A}'{B}'{C}'}}\Rightarrow {{V}_{A{B}'C{A}'{C}'}}=\dfrac{2}{3}\cdot \dfrac{{{a}^{3}}}{2}=\dfrac{{{a}^{3}}}{3}$