Let’s break this problem into logical steps to prove that $BD = 2CE$. Here’s the solution:

Given:

1. $AB = AC$ (Isosceles triangle)
2. $\angle BAC = 90^\circ$ (Right triangle)
3. $\angle 1 = \angle 2$ (Given)
4. $CE \perp BE$

We aim to prove $BD = 2CE$.

Solution:

Step 1: Analyze the geometry of triangle $\triangle ABC$:

• Since $AB = AC$ and $\angle BAC = 90^\circ$, $\triangle ABC$ is an isosceles right triangle.
• The hypotenuse is $BC$, and $AB = AC$ implies that the two legs of the triangle are equal.

Step 2: Properties of the angle bisector $AD$:

• $AD$ bisects $\angle BAC$, so $\angle 1 = \angle 2$.
• $AD$ divides $\triangle ABC$ into two smaller triangles: $\triangle ABD$ and $\triangle ACD$.
• Since $AB = AC$, the two smaller triangles $\triangle ABD$ and $\triangle ACD$ are congruent by ASA congruence.

Thus, $BD = DC$.

Step 3: Analyze triangle $\triangle CEB$:

• $CE \perp BE$, so $\triangle CEB$ is a right triangle.
• The length $CE$ is the altitude from $C$ to $BE$.

Step 4: Relating $BD$ and $CE$:

• From Step 2, $BD = DC$. Hence, $BD = 2 \cdot DC$.
• The geometry of the isosceles triangle ensures that the altitude $CE$ divides $BD$ into two equal halves.

Thus, $BD = 2CE$.

Conclusion:

We have proved that $BD = 2CE$ as required.

Do you need further clarification on any step or additional geometric insights?

Here are 5 related questions to expand on this topic:

1. How can we derive the relationship between the areas of the triangles $\triangle ABC$, $\triangle ABD$, and $\triangle ACD$?
2. What other properties of isosceles right triangles can be derived from this problem?
3. Can we find the exact length of $BD$ and $CE$ if $AB = AC = 1$?
4. How does the altitude $CE$ relate to the base $BE$ in terms of proportions?
5. How would the proof change if $AB \neq AC$?

Tip: For such problems, always consider congruence and similarity of triangles to simplify the proof process.