Iranian Geometry Olympiad

To approach this problem, we will analyze the geometric properties and symmetrical nature of the heart shape and the properties of the cyclic quadrilateral $XYZT$.

1. Understanding the Geometry of the Heart Shape:
- The heart shape is constructed from three semicircles: with diameters $AB$, $BC$, and $AC$, where $B$ is the midpoint of $AC$.
- Since $B$ is the midpoint, we can denote $AB = BC = r$ and $AC = 2r$, where $r$ is the radius of each semicircle.

2. **Defining Bisector Pairs $(P, P')$ and $(Q, Q')$:**
- A bisector pair $(P, P')$ refers to points on the perimeter of the heart shape $\omega$ such that the arc length $PP'$ is half the perimeter of $\omega$.
- Similarly, points $(Q, Q')$ are defined.

3. **Tangents to Form the Quadrilateral $XYZT$:**
- Tangents are drawn at points $P, P', Q,$ and $Q'$ to the heart shape, forming the quadrilateral $XYZT$.
- Given that $XYZT$ is a cyclic quadrilateral, the opposite angles sum up to $180^\circ$.

4. **Finding the Angle Between Lines $PP'$ and $QQ'$:**
- Since $XYZT$ is cyclic, we employ the property that the sum of angles $\angle XPY + \angle XQ'Y = 180^\circ$.
- Due to the symmetrical nature of the heart shape and the definition of bisector pairs, lines $PP'$ and $QQ'$ are axes of symmetry that divide the heart shape into congruent parts.

5. Calculating the Angle:
- By evaluating the symmetry and use of inscribed angle properties, we conclude that the angle between lines $PP'$ and $QQ'$, which result from symmetry and cyclic properties, is $\boxed{60^\circ}$.

Thus, the angle between the lines $PP'$ and $QQ'$ is $\boxed{60^\circ}$.