# 问题157: It is given that ABCD is a parallelogram, AM=MD and area of △CDN=12 . Find the area of the quadrilateral ABNM(x) .

It is given that $A B C D$ is a parallelogram, $A M=M D$ and area of $\triangle C D N=12$. Find the area of the quadrilateral $A B N M(x)$.

\begin{aligned} & \because \angle M N D=\angle C N B, \angle M D N=\angle C B N \therefore \triangle M N D \backsim \triangle C N B \\ & \because M \text { 是中点 } \therefore M D: A D=M D: C B=1: 2 \\ & \therefore M N: C N=M D: C B=1: 2 \\ & \because \triangle M N D \text { 与 } \triangle C N D \text { 等高且底边 } M N: C N=1: 2 \\ & \therefore S_{\triangle M N D}: S_{\triangle C N D}=1: 2 \\ & \therefore S_{\triangle M N D}=\frac{1}{2} S_{\triangle C N D}=\frac{1}{2} \cdot 12=6 \\ & \because S_{\triangle B N C}=\left(\frac{B C}{D M}\right)^2 \cdot S_{\triangle M N D}=2^2 \cdot 6=24 \\ & \therefore S_{\triangle C B D}=S_{\triangle C B N}+S_{\triangle C D N}=24+12=36 \\ & \therefore S_{\triangle A B D}=S_{\triangle C B D}=36 \\ & \therefore S_{A B N M}=S_{\triangle A B D}-S_{\triangle M N D}=36-6=30 \end{aligned}