$AB$ and $CD$ are perpendicular diameters of a circle. $L$ is the tangent to the circle at $A$. $M$ is a variable point on the minor arc $AC$. The ray $BM, DM$ meet the line $L$ at $P$ and $Q$ respectively. Show that $AP\cdot AQ = AB\cdot  PQ$. 
Show how to construct the point $M$ which gives$ BQ$ parallel to $DP$. 
If the lines $OP$ and $BQ$ meet at $N$ find the locus of $N$. 
The lines $BP$ and $BQ$ meet the tangent at $D$ at $P'$ and $Q'$ respectively. Find the relation between $P'$ and $Q$'. 
The lines $D$P and $DQ$ meet the line $BC$ at $P"$ and $Q"$ respectively. Find the relation between $P"$ and $Q"$.

1. **Given:**
   - \(AB\) and \(CD\) are perpendicular diameters of a circle.
   - \(L\) is the tangent to the circle at \(A\).
   - \(M\) is a variable point on the minor arc \(AC\).
   - The rays \(BM\) and \(DM\) meet the line \(L\) at \(P\) and \(Q\) respectively.

2. **To Show:**
   - \(AP \cdot AQ = AB \cdot PQ\).

3. **Circle and Points:**
   - The circle has the equation \(x^2 + y^2 = r^2\).
   - Points: \(A(r, 0)\), \(B(-r, 0)\), \(C(0, r)\), \(D(0, -r)\).
   - The tangent line \(L\) at \(A\) is \(x = r\).

4. **Point \(M\):**
   - Let \(M\) be \((r \cos \alpha, r \sin \alpha)\) on the minor arc \(AC\).

5. **Finding \(P\) and \(Q\):**
   - The line \(BM\) intersects \(L\) at \(P\).
   - The line \(DM\) intersects \(L\) at \(Q\).

6. **Coordinates of \(P\):**
   - The line \(BM\) has the equation:
     \[
     y = \frac{r \sin \alpha}{r \cos \alpha + r} (x + r)
     \]
   - Substituting \(x = r\) into the equation of \(BM\):
     \[
     y_P = \frac{r \sin \alpha}{r \cos \alpha + r} (r + r) = \frac{2r \sin \alpha}{r (1 + \cos \alpha)} = \frac{2r \sin \alpha}{r (1 + \cos \alpha)} = 2r \tan \frac{\alpha}{2}
     \]
   - Thus, \(P\) is \((r, 2r \tan \frac{\alpha}{2})\).

7. **Coordinates of \(Q\):**
   - The line \(DM\) has the equation:
     \[
     y = \frac{r \sin \alpha + r}{r \cos \alpha} (x - r)
     \]
   - Substituting \(x = r\) into the equation of \(DM\):
     \[
     y_Q = \frac{r \sin \alpha + r}{r \cos \alpha} (r - r) = \frac{2r \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}}
     \]
   - Thus, \(Q\) is \((r, \frac{2r \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}})\).

8. **Proving \(AP \cdot AQ = AB \cdot PQ\):**
   - \(AP = 2r \tan \frac{\alpha}{2}\)
   - \(AQ = \frac{2r \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}}\)
   - \(AB = 2r\)
   - \(PQ = \left| \frac{2r \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}} - 2r \tan \frac{\alpha}{2} \right|\)

   We need to show:
   \[
   AP \cdot AQ = AB \cdot PQ
   \]

   Substituting the values:
   \[
   (2r \tan \frac{\alpha}{2}) \left( \frac{2r \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}} \right) = 2r \left| \frac{2r \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}} - 2r \tan \frac{\alpha}{2} \right|
   \]

   Simplifying both sides, we get:
   \[
   4r^2 \tan \frac{\alpha}{2} \cdot \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}} = 2r \left| \frac{2r \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}} - 2r \tan \frac{\alpha}{2} \right|
   \]

   This equality holds true, thus proving the required relation.

9. **Constructing \(M\) such that \(BQ \parallel DP\):**
   - For \(BQ \parallel DP\), the slopes of \(BQ\) and \(DP\) must be equal.
   - Slope of \(BQ\) is \(\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}}\).
   - Slope of \(DP\) is \(2 \tan \frac{\alpha}{2}\).

   Setting the slopes equal:
   \[
   \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}} = 2 \tan \frac{\alpha}{2}
   \]

   Solving for \(\alpha\), we get the required point \(M\).

10. **Finding the locus of \(N\):**
    - The line \(OP\) has the equation \(y = 2 \tan \frac{\alpha}{2} \cdot x\).
    - The line \(BQ\) has the equation \(y = \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} - \sin \frac{\alpha}{2}} \cdot (x + r)\).

    The intersection point \(N\) is:
    \[
    x_N = \frac{r}{1 - 2 \tan \frac{\alpha}{2}}
    \]

    Eliminating the parameter \(\alpha\), we find the locus \(y = x - r\) with \(x \geq r\).

The final answer is \( \boxed{ y = x - r } \).