Let $\{P\}=A C \cap B D$ and let $\angle A P B=180-2 a$. Since $A B C D$ is an isosceles trapezoid, $A P B$ is an isosceles triangle. Therefore $\angle P B A=a$, which implies that $\angle P B I=90^{\circ}-a / 2$ since $I$ lies on the external bisector of $\angle P B A$. Since $I$ lies on the bisector of $\angle C P B$, it follows that $\angle B P I=a$ and hence that $I P B$ is isosceles with $|I P|=|P B|$. Similarly $J P C$ is isosceles with $|J P|=|P C|$. So, in the triangles $C P I$ and $B P J$ we have $P I \equiv P B$ and $P J \equiv C P$. Since $I$ and $J$ both lie on the internal bisector of $\angle B P C$, it follows that triangles $C P I$ and $B P J$ are congruent. Therefore $|I C|=|J B|$.

A competition of the Canadian Mathematical Society and supported by the Actuarial Profession.
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Expertise. Insight. Solutions.

1. Identify the properties of the trapezoid:
- Given $ABCD$ is a trapezoid with $AB \parallel CD$ and $|AB| > |CD|$.
- The edges $|AD| = |BC|$ are equal, indicating that $ABCD$ is an isosceles trapezoid.

2. Intersection of diagonals:
- Let $AC \cap BD = O$. In an isosceles trapezoid, the diagonals intersect at a point $O$ such that $\angle OAC = \angle OBD$ and $\angle OAD = \angle OBC$.

3. Angles at the intersection point:
- Since $AB \parallel CD$, the angles $\angle OAC = \angle OBD = \theta$ and $\angle OAD = \angle OBC = \theta$.

4. Centers of the circles:
- Let $I$ be the center of the circle tangent to lines $AB$, $AC$, and $BD$.
- Let $J$ be the center of the circle tangent to lines $CD$, $AC$, and $BD$.

5. Angles involving the centers:
- Since $I$ is the center of the circle tangent to $AB$, $AC$, and $BD$, the angle $\angle OAI = \frac{\theta}{2}$ because $AC$ is tangent to the circle with center $I$.
- Similarly, $\angle AOI = 180^\circ - \theta$ implies $\angle OIA = \frac{\theta}{2}$.

6. Equal distances from the intersection point:
- From the above, we have $OI = OB$ because $\angle OAI = \angle OIA$ and $O$ is the midpoint of $AC$ and $BD$ in terms of the tangency points.

7. **Similar argument for $J$:**
- By a similar argument, $OJ = OC$ because $\angle OAJ = \angle OJA$ and $O$ is the midpoint of $AC$ and $BD$ in terms of the tangency points.

8. Congruent triangles:
- We also have $\angle BOJ = \angle IOC$ because both are formed by the intersection of the diagonals and the tangency points.
- Combining $OI = OB$, $OJ = OC$, and $\angle BOJ = \angle IOC$, we get $\triangle BOJ \cong \triangle IOC$ by the Angle-Side-Angle (ASA) criterion.

9. Conclusion:
- Since $\triangle BOJ \cong \triangle IOC$, it follows that $|IC| = |JB|$.

$\blacksquare$

The final answer is $|IC| = |JB|$