Open WPF form when clicking WPF hyperlink

You could just handle the click event:

<Hyperlink Click="Hyperlink_Click">Link</Hyperlink>

private void Hyperlink_Click(object sender, RoutedEventArgs e)
{
    Dialogue diag = new Dialogue();
    diag.Show();
}

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You could also go crazy with XAML:

<Hyperlink>
    <Hyperlink.Style>
        <Style TargetType="{x:Type Hyperlink}">
            <Style.Triggers>
                <EventTrigger RoutedEvent="Hyperlink.Click">
                    <BeginStoryboard>
                        <Storyboard>
                            <ObjectAnimationUsingKeyFrames Storyboard.TargetProperty="Visibility">
                                <Storyboard.Target>
                                    <local:Dialogue />
                                </Storyboard.Target>
                                <DiscreteObjectKeyFrame KeyTime="0:0:0" Value="{x:Static Visibility.Visible}"/>
                            </ObjectAnimationUsingKeyFrames>
                        </Storyboard>
                    </BeginStoryboard>
                </EventTrigger>
            </Style.Triggers>
        </Style>
    </Hyperlink.Style>
    <Hyperlink.Inlines>
        <Run Text="Open Dialogue"/>
    </Hyperlink.Inlines>
</Hyperlink>

This however is very problematic since once the dialogue is closed it cannot be reopened, that means when you click the hypelink again an exception will be thrown.

---

Using interactivity you could do this without such problems (needs a reference to the Blend SDK though):

xmlns:i="clr-namespace:System.Windows.Interactivity;assembly=System.Windows.Interactivity"

<Hyperlink>
    <i:Interaction.Triggers>
        <i:EventTrigger EventName="Click">
            <t:CreateDialogAction Type="{x:Type local:Dialogue}"/>
        </i:EventTrigger>
    </i:Interaction.Triggers>
    <Hyperlink.Inlines>
        <Run Text="Open Dialogue"/>
    </Hyperlink.Inlines>
</Hyperlink>

The action for this:

public class CreateDialogAction : TriggerAction<Hyperlink>
{
    public Type Type { get; set; }

    protected override void Invoke(object parameter)
    {
        if (Type != null && Type.IsSubclassOf(typeof(Window)) && Type.GetConstructor(Type.EmptyTypes) != null)
        {
            Window window = Type.GetConstructor(Type.EmptyTypes).Invoke(null) as Window;
            window.Show();
        }
    }
}

You can achive this like this:

<Label Height="25" Margin="26,27,116,0" Name="label1" VerticalAlignment="Top">
    <Hyperlink Click="Hyperlink_Click">Click Me</Hyperlink>
</Label>

and handle it like this:

private void Hyperlink_Click(object sender, RoutedEventArgs e)
{
    Window2 form2 = new Window2();
    form2.Show();
}