A simpler way would be to add a new child when the open button is selected. If we do this we will not get an error, and we can get rid of the open menu item altogether.
We will combine the code for ShowNewForm and OpenFile.
Private Sub OpenFile(ByVal sender As Object, ByVal e As EventArgs) _
Handles OpenToolStripButton.Click
Dim ChildForm As New Form1
ChildForm.MdiParent = Me
m_ChildFormNumber += 1
ChildForm.Show()
Dim OpenFileDialog As New OpenFileDialog
OpenFileDialog.InitialDirectory = My.Computer.FileSystem.SpecialDirectories.MyDocuments
OpenFileDialog.Filter = _
"Picture Files (Bitmaps|*.bmp|GIFS|*.gif|JPG|*.jpg|All Files (*.*)|*.*"
If (OpenFileDialog.ShowDialog(Me) = System.Windows.Forms.DialogResult.OK) Then
Dim ActiveChild As Form1 = Me.ActiveMdiChild
ChildForm.Text = OpenFileDialog.FileName
ChildForm.Pic.Image = Image.FromFile(OpenFileDialog.FileName)
End If
End Sub 'OpenFile
MDINotepad