At the base of the pyramid $A B C D$ lies an isosceles right triangle $A B C$ with hypotenuse $A B=4$. The height of the pyramid is 2, and its base coincides with the midpoint of $A C$. Find the dihedral angle between the faces

$A B D$ and $A D C$.
Triangle $ABC$ is isosceles and right-angled, with its hypotenuse $AB$ equal to 4, so $AC = BC = 2\sqrt{2}$ and

$$
S_{\triangle ABC} = \frac{1}{2} \cdot (2\sqrt{2})^2 = 4
$$

Let $H$ be the midpoint of $AC$, and $K$ be the projection of point $H$ onto the hypotenuse $AB$. Then $DH = 2$, and $HK = 1$ as half the height $CL$ of triangle $ABC$. From the right triangles $DKH$ and $ADH$, we find that

$$
DK = \sqrt{HK^2 + DH^2} = \sqrt{1 + 4} = \sqrt{5}, \quad AD = \sqrt{AH^2 + DH^2} = \sqrt{2 + 4} = \sqrt{6}
$$

Then

$$
S_{\triangle ACD} = \frac{1}{2} AC \cdot DH = \frac{1}{2} \cdot 2\sqrt{2} \cdot 2 = 2\sqrt{2}
$$

From the theorem of three perpendiculars, it follows that $DK \perp AB$, so

$$
S_{\triangle ABD} = \frac{1}{2} AB \cdot DK = \frac{1}{2} \cdot 4 \cdot \sqrt{5} = 2\sqrt{5}
$$

We will calculate the volume of the pyramid in two ways. On one hand,

$$
V_{ABCD} = \frac{1}{3} S_{\triangle ABC} \cdot DH = \frac{1}{3} \cdot 4 \cdot 2 = \frac{8}{3}
$$

On the other hand, if $\alpha$ is the angle between the faces $ABD$ and $ACD$, then

$$
V_{ABCD} = \frac{2}{3} \cdot \frac{S_{\triangle ABD} \cdot S_{\triangle ACD} \sin \alpha}{AD} = \frac{2}{3} \cdot \frac{2\sqrt{5} \cdot 2\sqrt{2}}{\sqrt{6}} = \frac{8\sqrt{10}}{3\sqrt{6}} \sin \alpha
$$

From the equation

$$
\frac{8\sqrt{10}}{3\sqrt{6}} \sin \alpha = \frac{8}{3}
$$

we find that $\sin \alpha = \sqrt{\frac{3}{5}}$.

## Answer

$\arcsin \sqrt{\frac{3}{5}}$