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In parallelogram ABCD, M is the midpoint of CD.  
The area of triangle BNC is 42.  
Find the area of triangle MNC.

\(\text{Let the area of $\triangle ABN = A_1$} \\ \text{Let the area of $\triangle BNC = A_2=42$} \\ \text{Let the area of $\triangle MNC = A_3=\ ?$} \\ \text{Let $h=h_1+h_2$ or $h_2=h-h_1$} \)

\(\begin{array}{|rcll|} \hline A_3 &=& \dfrac{xh_1}{2} \\ \mathbf{2A_3} &=& \mathbf{xh_1} \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline A_1+A_2 &=& \dfrac{2xh}{2} \\ \mathbf{A_1+A_2} &=& \mathbf{xh} \qquad (2) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline A_1 &=& \dfrac{2xh_2}{2} \\ A_1 &=& xh_2 \quad | \quad h_2=h-h_1 \\ A_1 &=& x(h-h_1) \\ A_1 &=& xh-xh_1 \quad | \quad xh=A_1+A_2,\ xh_1=2A_3 \\ A_1 &=& A_1+A_2-2A_3 \\ 2A_3+A_1 &=& A_1+A_2 \\ 2A_3 &=& A_2 \\ A_3 &=& \dfrac{A_2}{2} \quad | \quad A_2 = 42 \\ A_3 &=& \dfrac{42}{2} \\ \mathbf{A_3} &=& \mathbf{21} \\ \hline \end{array}\)

The area of  \(\triangle MNC=\mathbf{21}\)