In the convex pentagon $ABCDE$: $\angle A = \angle C = 90^o$, $AB = AE, BC = CD, AC = 1$. Find the area of the pentagon.


1. **Constructing the auxiliary line and triangles:**
   - Let \( AK \) be a line perpendicular to \( AC \) such that \( AK = AC \). Since \( AC = 1 \), we have \( AK = 1 \).
   - Given \( AB = AE \) and \( \angle A = 90^\circ \), we know that \( \triangle CAB \) and \( \triangle KAE \) are right triangles with \( \angle CAK = \angle BAE \).

2. **Proving congruence of triangles:**
   - Since \( AB = AE \), \( AC = AK \), and \( \angle CAK = \angle BAE \), by the Angle-Side-Angle (ASA) criterion, \( \triangle CAB \cong \triangle KAE \).
   - Therefore, the areas of these triangles are equal: \([CAB] = [KAE]\).

3. **Analyzing the segments and angles:**
   - Given \( EK = BC = CD \), let \( \angle AKE = \theta \). Then, \( \angle EKC = 45^\circ - \theta \).
   - Since \( \angle BCA = \angle AKE = \theta \), it follows that \( \angle KCD = 45^\circ - \theta \).

4. **Parallel lines and congruent triangles:**
   - Since \( EK \parallel CD \), we have \( \angle KED = \angle CDE \).
   - Let \( KC \cap ED = P \). Then, \( \triangle CPD \cong \triangle KPE \) by the Angle-Side-Angle (ASA) criterion, implying \([CPD] = [KPE]\).

5. **Calculating the area of the pentagon:**
   - The area of the pentagon \( ABCDE \) can be decomposed as follows:
     \[
     [ABCDE] = [ABC] + [AEPC] + [CPD]
     \]
   - Since \( [ABC] = [AEK] \) and \( [CPD] = [KPE] \), we have:
     \[
     [ABCDE] = [AEK] + [AEPC] + [KPE]
     \]
   - Notice that \( [AEK] + [KPE] = [AKC] \), and since \( AK = AC = 1 \), the area of \( \triangle AKC \) is:
     \[
     [AKC] = \frac{1}{2} \times AC \times AK = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}
     \]

The final answer is \(\boxed{\frac{1}{2}}\).