In a triangular pyramid $A B C D$, find the angle between the lines $A D$ and $B C$, if $A B=A C$ and $\angle D A B=\angle D A C$.
Let $DO$ be the height of the pyramid, and points $M$ and $N$ be the bases of the perpendiculars dropped from point $O$ to lines $AB$ and $AC$, respectively. According to the theorem of three perpendiculars, $DM \perp AB$ and $DN \perp AC$. Right triangles $DAM$ and $DAN$ are equal by hypotenuse and acute angle, so $OM = ON$, i.e., point $O$ is equidistant from the sides of angle $CAB$. Therefore, point $O$ lies on the bisector $AK$ of the isosceles triangle $ABC$ (or its extension). Since $AK \perp BC$ and $AK$ is the orthogonal projection of the inclined line $AD$ onto the plane of the base $ABC$, by the theorem of three perpendiculars, $AD \perp BC$. Thus, the angle between lines $AD$ and $BC$ is $90^\circ$.

## Answer

Prove that through a point not lying on a plane, one can draw a unique plane parallel to the given one.

## Solution

Through an arbitrary point of the given plane $\alpha$, draw two intersecting lines $a$ and $b$ (Fig.1). Through the given point $M$, not lying in the plane $\alpha$, draw lines $a1$ and $b1$, respectively parallel to lines $a$ and $b$. Through the intersecting lines $a1$ and $b1$, draw the plane $\beta$. By the criterion for parallel planes, planes $\alpha$ and $\beta$ are parallel. The existence of the plane is proven. Now we will prove that such a plane is unique. Suppose that through the point $M$, not lying in the plane $\alpha$, one can draw at least two planes $\beta$ and $\beta1$, parallel to the plane $\alpha$ (Fig.2). Let $c$ be the line of intersection of planes $\beta$ and $\beta1$. Take a line $d$ on the plane $\alpha$ that is not parallel to the line $c$. Through the point $M$ and the line $d$, draw the plane $\gamma$. This plane intersects planes $\beta$ and $\beta1$ along lines parallel to the line $d$. Thus, through the point $M$ in the plane $\gamma$, there are two lines parallel to the line $d$ in the plane $\gamma$, which is impossible.