Access base class protected methods in static block

Nadie ha tomado este issue todavía.

### 🔍 Search Terms

- Dynamic method override
- Static block super
- Static block prototype

### ✅ Viability Checklist

- [x] This wouldn't be a breaking change in existing TypeScript/JavaScript code
- [x] This wouldn't change the runtime behavior of existing JavaScript code
- [x] This could be implemented without emitting different JS based on the types of the expressions
- [x] This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, new syntax sugar for JS, etc.)
- [x] This isn't a request to add a new utility type: https://github.com/microsoft/TypeScript/wiki/No-New-Utility-Types
- [x] This feature would agree with the rest of our Design Goals: https://github.com/Microsoft/TypeScript/wiki/TypeScript-Design-Goals

### ⭐ Suggestion

Hey there, I have a use case where I want to declare/override a protected method using it's prototype.

With the addition of static blocks I can now override the method itself, however I can't call the same method of the base class using super. This makes it impossible to create dynamic overrides of a prototype method atm.

My suggestion would be to allow the usage of the base type prototypes in a static block of the class as well.
This would make it a lot easier to dynamically override/declare a method without using instance fields.

### 📃 Motivating Example

TypeScript now supports dynamic method overrides using static initialization blocks instead of runtime checks.

### 💻 Use Cases

**1. What do you want to use this for?**
Dynamic method declarations / overrides
Dynamic class creation using an options object.

**2. What shortcomings exist with current approaches?**
Currently the only way I see to do this is either:
- Use as any/ which destroys type information
- Use as DerivedType which allows the call with type information
- Always override the method and call the additional method in an if branch which usually requires an instance field
- Declare the method on an instance field instead of the prototype which uses more memory
- Use @ts-ignore because you know what you are doing

**3. What workarounds are you using in the meantime?**
- I use (Base.prototype as DerivedType).myMethod which imho is the best solution atm

**Example:**
https://www.typescriptlang.org/play/?ssl=25&ssc=2&pln=1&pc=1#code/MYGwhgzhAECyCeBhcVoG8BQ1oAcBOA9gC4CmwpAJtALbywlEAWBFAFAJTpbbQD0v0AHTCIBaiWggCAcwCWwbgF8MyjKEgwEAERJ5ZANxIVkG6CQAepAHYVNSFBEzZ8xMpWgFDePRQm16TCwcXDx8AsKC0KLi0GAUFLJEsgRWYCCSMvLc2BAArji6gv4MzGzsShjcEERgScAhPEyyEIIuRMTwBUV0JSzQALzQAGa5VuTJVqxNEABccPA6eobGDpxOoWFCwlFiEnEJSSlpGXIK2Tx5BXjdAaXB-FH5utCjvup4RtBEEABMAGx-AAMlQ28xMUFahHaRE6JBuvQogmAaRAU0YzXYAG5NlZiLEQFIAO6fb4-AAsZL+II2rAQ4JabQ6BViMDAVng7HhgURyIJaIxmwAKuiYISCHgANYwABGuSI0GaLysEDAQxI52wtPsGkhxCZezsiwMRnpnOK3KRKP5EE4D2FirFkplcoVMFGKrV52U2GUiiAA

Comienza con el ejemplo enlazado de TypeScript Playground y la discusión del issue sobre bloques estáticos, métodos protegidos y sobrescrituras de prototipos. Sigue cómo se comprobaría el acceso al prototipo base propuesto mediante el sistema de tipos. Se considera terminado cuando el comportamiento está claramente definido y se cubren los casos relevantes de comprobación de tipos.

Escrito por el modelo de indexación a partir del texto del issue.