46. In triangle $ABC$, angle $\hat{B}$ is $\pi / 4$, and angle $\hat{C}$ is $\pi / 6$. Circles are constructed on medians $BM$ and $CN$ as diameters, intersecting at points $P$ and $Q$. Chord $PQ$ intersects side $BC$ at point $D$. Find the ratio $|BC|:|DC|$.
46. Note (Fig. 8) that \( P Q \perp C B \). Let \( T \) be the intersection point of \( M N \) and \( P Q \), and \( L \) and \( K \) be the feet of the perpendiculars dropped from \( C \) and \( B \) to the line \( M N \) ( \( L \) and \( K \) lie on the circles constructed with \( C N \) and \( B M \) as diameters). Using the properties of intersecting chords in circles, we get

\[
\begin{aligned}
& |P T| \cdot |T Q| = |N T| \cdot |L T|, \\
& |P T| \cdot |T Q| = |M T| \cdot |T K| \cdot
\end{aligned}
\]

But \( |L T| = |C D|, |T K| = |D B| \) (since \( C L K B \) is a rectangle, and \( P Q \perp C B \)). Thus, \( |N T| \cdot |C D| = |M T| \cdot |D B|, \frac{|M T|}{|N T|} = \frac{|C D|}{|D B|} \), i.e., the line \( P Q \) divides \( C B \) and \( M N \) in the same ratio; hence, \( P Q \) passes through \( A \), and \( D \) is the foot of the altitude.

Answer: \(\frac{|B D|}{|D C|} = \frac{1}{\sqrt{3}}\).

![](https://cdn.mathpix.com/cropped/2024_05_21_6282cf4a54d2a9633c48g-075.jpg?height=371&width=417&top_left_y=469&top_left_x=197)

Fig. 8.

![](https://cdn.mathpix.com/cropped/2024_05_21_6282cf4a54d2a9633c48g-075.jpg?height=365&width=400&top_left_y=472&top_left_x=640)

Fig. 9.