In the triangular pyramid $A B C D$, the edges $A B$ and $C D$ are mutually perpendicular, $A D = B C$, the distance from the midpoint $E$ of edge $A B$ to the plane $A C D$ is $h$, $\angle D A C = \frac{\pi}{2}$, $\angle A C D = \frac{\pi}{4}$, and the angle between edge $D C$ and the face $A B C$ is $\frac{\pi}{6}$. Find the distance from point $E$ to the plane $B C D$, the angle between edge $A B$ and the face $A C D$, and the angle between the faces $A B D$ and $A B C$.
Let $D H, A P$ and $B F$ be the heights of the pyramid $A B C D$, and $G$ be the orthogonal projection of point $E$ onto the plane of face $A D C$. Since $A F$ is the orthogonal projection of the inclined line $A B$ onto the plane $A D C$, points $A, G$ and $F$ lie on the same line, and $E G$ is the midline of the right triangle $A B F$, so $B F=2 E G=2 h$. Similarly, we can prove that the height $A P$ of the pyramid is twice the desired distance from point $E$ to the plane $B C D$.

Let $A D=B C=a$. From the isosceles right triangle $A D C$, we find that $A C=a$ and $D C=a \sqrt{2}$. From the problem statement, it follows that the height $D H$ is the leg of the right triangle $D C H$ opposite the angle of

$$
\begin{gathered}
30^{\circ}, \text { therefore } \\
D H=\frac{1}{2} D C=\frac{1}{2} \cdot a \sqrt{2}=\frac{a \sqrt{2}}{2}, C H=D H \sqrt{3}=\frac{a \sqrt{6}}{2} . \\
\text { Then } \\
A H=\sqrt{A D^{2}-D H^{2}}=\sqrt{a^{2}-\frac{a^{2}}{2}}=\frac{a \sqrt{2}}{2} .
\end{gathered}
$$

By the theorem of three perpendiculars, $C H \perp A B$, and since triangle $A C B$ is isosceles ($C B=A C=a$), its height lying on the line $C H$ passes through the midpoint $E$ of the base $A B$. In triangle $A D B$, the median $D E$ is also an altitude ($D E \perp A B$ by the theorem of three perpendiculars), so $D B=D A=a$. Therefore, triangle $D B C$, which is congruent to triangle $D A C$ by three sides, is also a right and isosceles triangle. By the cosine theorem,

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-47.jpg?height=131&width=832&top_left_y=2770&top_left_x=613)

$$
C E=A C \cos A C H=a \sqrt{\frac{2}{3}}, A E=A C \sin \angle A C H=a \cdot \sqrt{1-\frac{2}{3}}=\frac{a}{\sqrt{3}} \text {, }
$$

Writing the volume of the pyramid $A B C D$ in two ways ($\left.\frac{1}{3} S_{\triangle B C D} \cdot A P=\frac{1}{3} S_{\triangle A C D} \cdot B F\right)$, we find that the height $A P$ is equal to the height $B F$. Therefore, the distance from point $E$ to the plane $B C D$ is equal to the distance from point $E$ to the plane $A C D$, i.e., $h$. On the other hand, $\frac{1}{\overline{3}} S_{\triangle A C D} \cdot B F=\frac{1}{3} S_{\triangle A B C} \cdot D H$, or

$$
\frac{1}{3} \cdot \frac{1}{2} a 2 \cdot 2 h=\frac{1}{3} \cdot \frac{a}{\sqrt{3}} \cdot a \sqrt{\frac{2}{3}} \cdot \frac{a \sqrt{2}}{2}
$$

from which $a=3 h$. Let $\alpha$ be the desired angle between the edge $A B$ and the face $A C D$. From the right triangle $A G E$, we find that

$$
\sin \alpha=\sin \angle E A G=\frac{E G}{A E}=\frac{\frac{h}{\frac{a}{3}}}{\sqrt{3}}=\frac{\frac{h}{\sqrt[3]{3}}}{\sqrt{3}}=\frac{1}{\sqrt{3}}
$$

Since $D E$ and $C E$ are the heights of the equal isosceles triangles $A D B$ and $A B C$, dropped to the common base $A B$, the linear angle of the dihedral angle between the faces $A D B$ and $A B C$ is the angle $D E C$. By the cosine theorem,

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-48.jpg?height=132&width=820&top_left_y=926&top_left_x=619)

$$
\begin{aligned}
& \text { Therefore, } \angle D E C=120^{\circ} \text {. }
\end{aligned}
$$

## Answer

## $h, \arcsin \frac{1}{\sqrt{3}}, \frac{2 \pi}{3}$.  Submit a comment

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-48.jpg?height=183&width=2041&top_left_y=1438&top_left_x=0)

In the triangular pyramid $A B C D$, the edges $A C$ and $B D$ are mutually perpendicular, $A B=B D=A D=a$, the midpoint of edge $A C$ is equidistant from the planes $A B D$ and $B C D$, the angle between the edge $A C$ and the face $C B D$ is $\arcsin \frac{1}{\sqrt{3}}$. Find the edge $C D$, the angle $C A D$, and the angle between the edge $B D$ and the face $A C D$.

## Answer

$$
a, \frac{\pi}{4}, \frac{\pi}{4}
$$

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