Solid Geometry Advanced – Problem 4: Prove that the two lateral faces plane and plane are perpendicular

In a right triangular prism $ABC-A_1B_1C_1$, the lateral edges are perpendicular to the base plane, and $\angle ABC=90^\circ$.

Prove that the two lateral faces plane $ABB_1A_1$ and plane $BCC_1B_1$ are perpendicular.

(1) Because it is a right prism, $BB_1\perp$ plane $ABC$. Hence $BB_1\perp AB$ and $BB_1\perp BC$.

(2) Given $\angle ABC=90^\circ$, we have $AB\perp BC$.

(3) In plane $BCC_1B_1$, the two intersecting lines are $BC$ and $BB_1$. From (1)–(2), line $AB$ is perpendicular to both $BC$ and $BB_1$, so $$AB\perp \text{plane }BCC_1B_1.$$ (4) Since $AB\subset$ plane $ABB_1A_1$, plane $ABB_1A_1$ contains a line perpendicular to plane $BCC_1B_1$. Therefore $$\text{plane }ABB_1A_1\perp \text{plane }BCC_1B_1.$$

Plane $ABB_1A_1$ is perpendicular to plane $BCC_1B_1$.