Why does my program stop working before an if statement? [duplicate]

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• How do I compare strings in Java? 23 answers

i'm working on a random program for homothety. when i run it, my program stops before an if-statement; first i tought it was a bug with the "nextint()" so i added lines, but i saw that there is an problem with my if-statement.

could anybody help me with this bug? PS: sorry, i'm from belgium so the Strings are typed in Dutch, and sorry for my rediculas spelling of words in English.

//vars
    int x;
    int y;
    int z;
    String categ;
    Scanner i = new Scanner(System.in);
    int k;
    int xres;
    int yres;
    int zres;
    //input
    System.out.println("geef de vorm van het originele voorwerp als volgt op:");
    System.out.println("A = driehoek");
    System.out.println("B = vierkant");
    categ = i.nextLine();

    if(categ == "A" || categ == "a"){

        System.out.println("geef zijde x in: ");
        x = i.nextInt();
        System.out.println("geef zijde y in: ");
        y = i.nextInt();
        System.out.println("geef zijde z in: ");
        z = i.nextInt();
        System.out.println("geef de vergrotingsfactor in: ");
        k = i.nextInt();

        xres = x * k;
        yres = y * k;
        zres = z * k;

        System.out.println("\nna homothetie is\nx " 
        + xres +" cm,\ny: " 
                + yres + " cm\nen z: " 
        + zres + " cm.");

    }else{
        if(categ == "B" || categ == "b"){

            System.out.println("geef zijde x in: ");
            x = i.nextInt();
            System.out.println("geef zijde y in: ");
            y = i.nextInt();
            System.out.println("geef de vergrotingsfactor in: ");
            k = i.nextInt();

            xres = x * k;
            yres = y * k;

            System.out.println("\nna homothetie is\nx " 
                    + xres +" cm,\nen y " 
                            + yres + " cm");`

...