If BM : MD = 1 : 4, then [tex]BM=x\ in,\ MD=4x\ in.[/tex] Triangles AMD and CMB are similar, then
[tex]\dfrac{BM}{MD}=\dfrac{CM}{MA}=\dfrac{1}{4}.[/tex]
Let [tex]CM=y\ in,[/tex] then [tex]MA=4y\ in.[/tex]
Given [tex]A_{\triangle ABM}=8\ in^2.[/tex] Note that
[tex]A_{\triangle ABM}=\dfrac{1}{2}\cdot BM\cdot MA\cdot \sin\alpha=\dfrac{1}{2}\cdot x\cdot 4y\cdot \sin\alpha=8,\\ \\xy\sin\alpha=4.[/tex]
1. Consider triangle CMD. The area of this triangle is
[tex]A_{\triangle CMD}=\dfrac{1}{2}CM\cdot MD\cdot \sin\alpha=\dfrac{1}{2}\cdot y\cdot 4x\cdot \sin\alpha=2xy\sin\alpha=2\cdot 4=8\ in^2.[/tex]
2. Consider triangle AMD. The area of this triangle is
[tex]A_{\triangle AMD}=\dfrac{1}{2}\cdot AM\cdot MD\cdot \sin(180^{\circ}-\alpha)=\dfrac{1}{2}\cdot 4y\cdot 4x\cdot \sin\alpha=8xy\sin\alpha=8\cdot 4=32\ in^2.[/tex]
3. Consider triangle ACD. The area of this triangle is
[tex]A_{\triangle ACD}=A_{\triangle AMD}+A_{\triangle CMD}=32+8=40\ in^2.[/tex]
Answer: [tex]A_{\triangle CMD}=8\ in^2,\ A_{\triangle AMD}=32\ in^2,\ A_{\triangle ACD}=40\ in^2.[/tex]
