100 points and !Images to help solve the stuff please I need it to stay in school

Will give brainliest! 2017 amc 12a problem 24 quadrilateral [tex]abcd[/tex] is inscribed in circle [tex]o[/tex] and has side lengths [tex]ab=3, bc=2, cd=6[/tex], and [tex]da=8[/tex]. let [tex]x[/tex] and [tex]y[/tex] be points on [tex]\overline{bd}[/tex] such that [tex]\frac{dx}{bd} = \frac{1}{4}[/tex] and [tex]\frac{by}{bd} = \frac{11}{36}[/tex]. let [tex]e[/tex] be the intersection of line [tex]ax[/tex] and the line through [tex]y[/tex] parallel to [tex]\overline{ad}[/tex]. let [tex]f[/tex] be the intersection of line [tex]cx[/tex] and the line through [tex]e[/tex] parallel to [tex]\overline{ac}[/tex]. let [tex]g[/tex] be the point on circle [tex]o[/tex] other than [tex]c[/tex] that lies on line [tex]cx[/tex]. what is [tex]xf\cdot xg[/tex]? show all work!

1. find the area of the unshaded square 2. find the area of the large square 3. what is the area of the frame represented by the shaded region show

If m∠abc = 70°, what is m∠abd? justify your reasoning.  using the addition property of equality, 40 + 70 = 110, so m∠abd = 110°.  using the subtraction property of equality, 70 − 30 = 40, so m∠abd = 30°.  using the angle addition postulate, 40 + m∠abd = 70. so, m∠abd = 30° using the subtraction property of equality.  using the angle addition postulate, 40 + 70 = m∠abd. so, m∠abd = 110° using the addition property of equality.