How to bold highlight duplicate rows excel vba - excel

Question

I have the following code which picks up duplicate rows, however I can not get the code to highlight the duplicates in bold as well as deleting them at the same time.
Public Sub ProcessData()
Const TEST_COLUMN As String = "A" '<=== change to suit
Dim i As Long
Dim iLastRow As Long
Dim rng As Range
With ActiveSheet
iLastRow = .Cells(.Rows.Count, TEST_COLUMN).End(xlUp).Row
For i = 1 To iLastRow
If .Evaluate("SUMPRODUCT(--(A" & i & ":A" & iLastRow & "=A" & i & ")," & _
"--(D" & i & ":D" & iLastRow & "=D" & i & ")," & _
"--(F" & i & ":F" & iLastRow & "=F" & i & ")," & _
"--(J" & i & ":J" & iLastRow & "=J" & i & ")," & _
"--(K" & i & ":K" & iLastRow & "=K" & i & "))") > 1 Then
If rng Is Nothing Then
Set rng = .Cells(i, "A").Resize(, 11)
Else
Set rng = Union(rng, .Cells(i, "A").Resize(, 11))
End If
End If
Next i
**If Not rng Is Nothing Then rng.Delete.font.bold = true**
End With
End Sub
the example of the dataset and desired output can be seen in the following downloadable link below:
https://www.dropbox.com/s/7rhktg6b4nk6ig0/Bold_highlight_Duplicate%20.xlsm
any help would be very much appreciated. Thank you.
Edit:
to clarify, this is how it should look like, just that input shall be deleted and the bold highlighting should appear in the output section:

Answer 1

Instead of **If Not rng Is Nothing Then rng.Delete.font.bold = true** use the following:
If Not rng Is Nothing Then
with rng
.Offset(.Areas(.Areas.Count).Rows(.Areas(.Areas.Count).Rows.Count).Row + 1).Font.Bold = True
.Delete
end with
End If
How this works?
Well, you could have setup the bold indicator while testing for duplicates, however you took a different approach, which does not allow that.
So, your rng is a multiarea selection.
You have to get to the last area, then to the last row of that area, and then retrieve the actual row you're in. Then add +1 for the space between. Now you know how many rows are covered by the input section + the gap to the output and you offset your selection by this count onto the output section.
However, there might be complications, depenting on your input/output - I tested this briefly on your example - worked. Still, I think it would be better to use a different kind of loop & duplicate detection.