Дано:

\(ABCD\) — параллелограмм;

\(\angle\) \(BCA =\) 23\(\degree\);

\(\angle\) \(BAC =\) 17\(\degree\).

Найти:

\(\angle\) \(BAD =\) 40\(\degree\);        \(\angle\) \(B =\) 140\(\degree\);

\(\angle\) \(BCD =\) 40\(\degree\);        \(\angle\) \(D =\) 140\(\degree\).