Let the lateral edge $A A_1$ of the given triangular prism $A B C A_1 B_1 C_1$ form angles of $45^{\circ}$ with the sides $A B$ and $A C$. Drop a perpendicular $A_1 P$ from vertex $A_1$ to the plane of the base $A B C$, and from point $P$ drop perpendiculars $P M$ and $P N$ to the lines $A B$ and $A C$. By the theorem of three perpendiculars, $A_1 M \perp A B$ and $A_1 N \perp A C$. The right triangles $A A_1 M$ and $A A_1 N$ are equal by hypotenuse and acute angle, so $P M = P N$, i.e., point $P$ is equidistant from the sides $A B$ and $A C$ of angle $B A C$. Therefore, point $P$ lies on the bisector of this angle, and since the bisector of an isosceles triangle is its altitude, $A P \perp B C$. Thus, the orthogonal projection of the inclined line $A A_1$ onto the plane of the base is perpendicular to the line $B C$. By the theorem of three perpendiculars, $A A_1 \perp B C$. Therefore, $B B_1 \perp B C$ and $C C_1 \perp B C$, and the parallelogram $B B_1 C_1 C$ is a rectangle. Further, we have:

$$
S_{B B_1 C_1 C} = B C \cdot B B_1 = a b, \quad A_1 N = A_1 M = A A_1 \sin \angle A_1 A M = \frac{b \sqrt{2}}{2}
$$

$$
S_{A A_1 C_1 C} = S_{A A_1 B_1 B} = A B \cdot A_1 M = \frac{a b \sqrt{2}}{2}.
$$

If $S$ is the lateral surface area of the prism, then

$$
\begin{gathered}
S = S_{B B_1 C_1 C} + S_{A A_1 C_1 C} + S_{A A_1 B_1 B} = \\
= S_{B B_1 C_1 C} + 2 S_{A A_1 B_1 B} = a b + a b \sqrt{2} = a b (1 + \sqrt{2})
\end{gathered}
$$

## Answer

$a b (1 + \sqrt{2})$.
8,9

The base of an oblique prism is an equilateral triangle with side $a$. One of the lateral edges is equal to $b$ and forms angles of $45^{\circ}$ with the adjacent sides of the base. Find the lateral surface area of the prism.