For an isosceles triangle $ABC$ where $AB=AC$, it is possible to construct, using only compass and straightedge, an isosceles triangle $PQR$ where $PQ=PR$ such that triangle $PQR$ is similar to triangle $ABC$, point $P$ is in the interior of line segment $AC$, point $Q$ is in the interior of line segment $AB$, and point $R$ is in the interior of line segment $BC$. Describe one method of performing such a construction. Your method should work on every isosceles triangle $ABC$, except that you may choose an upper limit or lower limit on the size of angle $BAC$.
[asy]
defaultpen(linewidth(0.7));
pair a= (79,164),b=(19,22),c=(138,22),p=(109,91),q=(38,67),r=(78,22);
pair point = ((p.x+q.x+r.x)/3,(p.y+q.y+r.y)/3);
draw(a--b--c--cycle);
draw(p--q--r--cycle);
label("$A$",a,dir(point--a));
label("$B$",b,dir(point--b));
label("$C$",c,dir(point--c));
label("$P$",p,dir(point--p));
label("$Q$",q,dir(point--q));
label("$R$",r,dir(point--r));[/asy]
1. **Identify the Midpoints**: 
   - Let \( M \) be the midpoint of \( AC \).
   - Let \( N \) be the midpoint of \( BC \).

2. **Construct the Medial Triangle**:
   - The medial triangle of \( \triangle ABC \) is formed by connecting the midpoints of the sides of \( \triangle ABC \). This triangle is similar to \( \triangle ABC \) and has half the side lengths.
   - In this case, the medial triangle will have vertices at \( M \), \( N \), and the midpoint of \( AB \), which we will call \( L \).

3. **Construct the Circle**:
   - Draw a circle with center \( M \) and radius \( MN \). This circle will intersect \( AB \) at a point \( Q \) such that \( Q \) is in the interior of \( AB \).

4. **Identify Point \( P \)**:
   - Point \( P \) is already identified as the midpoint \( M \) of \( AC \).

5. **Identify Point \( R \)**:
   - Point \( R \) is already identified as the midpoint \( N \) of \( BC \).

6. **Verify Similarity**:
   - Since \( \triangle MNL \) is the medial triangle of \( \triangle ABC \), it is similar to \( \triangle ABC \) by the Midline Theorem, which states that the segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long.

7. **Check the Conditions**:
   - Point \( P \) (which is \( M \)) is in the interior of \( AC \).
   - Point \( Q \) is in the interior of \( AB \).
   - Point \( R \) (which is \( N \)) is in the interior of \( BC \).

Thus, the construction of \( \triangle PQR \) where \( P = M \), \( Q \) is the intersection of the circle centered at \( M \) with radius \( MN \) and \( AB \), and \( R = N \) satisfies all the given conditions.

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