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

Đặt .
Ta có: $\left\{ \begin{aligned}
& M\in \left( MBC \right)\cap \left( M{B}'{C}' \right) \\
& BC\subset \left( MBC \right);{B}'{C}'\subset \left( M{B}'{C}' \right) \\
& BC//{B}'{C}' \\
\end{aligned} \right.$$\Rightarrow \left( MBC \right)\cap \left( M{B}'{C}' \right)=\Delta \Delta M \Delta //BC//{B}'{C}'I,JBC{B}'{C}'MI\bot BCMJ\bot {B}'{C}'MBCM{B}'{C}'MMI\bot \Delta MJ\bot \Delta \left\{ \begin{aligned}
& \left( MBC \right)\cap \left( M{B}'{C}' \right)=\Delta \\
& MI\subset \left( MBC \right),MI\bot \Delta \\
& MJ\subset \left( M{B}'{C}' \right),MJ\bot \Delta \\
\end{aligned} \right.\Rightarrow \left( (MBC);(M{B}'{C}') \right)=\left( MI;MJ \right)=90{}^\circ AB=AC=\dfrac{a}{\sqrt{2}};AI=\dfrac{a}{2}MI=MJ=\sqrt{M{{A}^{2}}+A{{I}^{2}}}=\sqrt{\dfrac{{{h}^{2}}}{4}+\dfrac{{{a}^{2}}}{4}}MIJMI{{J}^{2}}=2M{{I}^{2}}\Leftrightarrow {{h}^{2}}=\dfrac{{{h}^{2}}}{2}+\dfrac{{{a}^{2}}}{2}\Rightarrow h=aABC.{A}'{B}'{C}'{{V}_{ABC.{A}'{B}'{C}'}}={{S}_{ABC}}.A{A}'=\dfrac{1}{2}.AB.AC.A{A}'=\dfrac{1}{2}.\dfrac{a}{\sqrt{2}}.\dfrac{a}{\sqrt{2}}.a=\dfrac{{{a}^{3}}}{4}$.