How can I open an image

Question

I want to open an image and set it as Source of Image_1.

Next is code , but I don't know how to use CreateStreamedFileAsync. Anyone can help me.

Windows::Storage::StreamedFileDataRequestedHandler^ dataRequested;
Windows::Storage::Streams::IRandomAccessStreamReference^ thumbnail;
IAsyncOperation<Windows::Storage::StorageFile^>^ fileAsync = StorageFile::CreateStreamedFileAsync(TEXT("Assets/win8.png"), dataRequested, thumbnail);
	
auto imageTask = create_task(fileAsync);
imageTask.then([this](StorageFile^ file)
{
	if (file)
	{
		// Ensure the stream is disposed once the image is loaded
		create_task(file->OpenAsync(Windows::Storage::FileAccessMode::Read)).then([this](IRandomAccessStream^ fileStream)
		{
		// Set the image source to the selected bitmap
		auto bitmapImage = ref new BitmapImage();
		bitmapImage->DecodePixelHeight = 100;
		bitmapImage->DecodePixelWidth = 100;

		bitmapImage->SetSource(fileStream);
		Image_1->Source = bitmapImage;
	});
                 }
});