Let me analyze the content of the uploaded image.

Problems Translation and Interpretation:

Bài 1:

Given a triangle $\Delta ABC$ with $\angle A = 90^\circ$, let $M$ and $N$ be the midpoints of $AC$ and $AB$, respectively.
a) Prove that $BM = CN$.
b) Let $I$ be the intersection of $BM$ and $CN$. Prove:

• $\Delta IBN = \Delta ICM$
• $\Delta IBC$ is isosceles.

Bài 2:

For the given diagram:

• $\angle ADB = \angle BCA = 90^\circ$, $AD = BC$.
  Prove:
  a) $\Delta ADI = \Delta BCI$.
  b) $AB \parallel CD$.

Let me solve these problems step by step.

Solution:

Bài 1:

a) Prove $BM = CN$:

1. Since $M$ and $N$ are midpoints, we have:

   • $AM = MC = \frac{1}{2}AC$
   • $AN = NB = \frac{1}{2}AB$

2. In $\Delta ABC$ (right triangle):

   • By the Pythagorean theorem:
     $AB^2 + AC^2 = BC^2$.

3. Using the midpoint theorem:
   The medians $BM$ and $CN$ divide $\Delta ABC$ into congruent halves:

   • $BM$ and $CN$ are symmetric, and since $\Delta ABC$ is right-angled, $BM = CN$.

b) Prove $\Delta IBN = \Delta ICM$:

1. Since $I$ is the intersection point of $BM$ and $CN$, we consider the triangles $\Delta IBN$ and $\Delta ICM$.

2. Both triangles share:

   • $\angle IBN = \angle ICM$ (opposite angles at $I$).
   • $BN = CM$ (from part a).
   • $BI = CI$ (since $I$ is the intersection point, and medians divide the triangles symmetrically).

3. By the SAS criterion, $\Delta IBN = \Delta ICM$.

Prove $\Delta IBC$ is isosceles:

1. In $\Delta IBC$, from part b:
   • $BI = CI$, so $\Delta IBC$ is isoscel