[ Perpendicularity of a line and a plane (other).]

In the pyramid $A B C D$, the edges $A D, B D$ and $C D$ are equal to 5, the distance from point $D$ to the plane $A B C$ is 4. Find the radius of the circle circumscribed around triangle $A B C$.
Let $R$ be the radius of the circle circumscribed around triangle $ABC$. Since the lateral edges of the pyramid $ABCD$ with vertex $D$ are equal, the height $DO$ of the pyramid passes through the center $O$ of the circle circumscribed around the base $ABC$. The line $DO$ is perpendicular to the plane of the base $ABC$, and therefore it is perpendicular to the line $OA$. From the right triangle $AOD$, we find that

$$
OA = \sqrt{DA^2 - DO^2} = \sqrt{25 - 16} = 3
$$

## Answer

Prove that if a line $p$ forms equal angles with three pairwise intersecting lines in a plane, then the line $p$ is perpendicular to this plane.

## Solution

Let the line $p$ intersect the plane $\alpha$, containing the given lines $k, m$, and $n$, at point $O$. Draw through point $O$ lines $k_1, m_1$, and $n_1$, respectively parallel to lines $k, m$, and $n$. Then the line $p$ forms equal angles with lines $k_1, m_1$, and $n_1$. Therefore, any point $A$ on the line $p$, different from $O$, is equidistant from these lines. Hence, the orthogonal projection $A_1$ of point $A$ onto the plane $\alpha$ is also equidistant from lines $k_1, m_1$, and $n_1$. Therefore, point $A_1$ lies on the bisector of one of the angles formed by lines $k_1$ and $m_1$, on the bisector of one of the angles formed by lines $k_1$ and $n_1$, and on the bisector of one of the angles formed by lines $m_1$ and $n_1$. Since lines $k_1, m_1$, and $n_1$ are distinct, the specified bisectors have only one common point $O$. Therefore, point $A_1$ coincides with point $O$. Then the line $p$ coincides with the line $AA_1$.

Therefore, the line $p$ is perpendicular to the plane $\alpha$.