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Mathematics, 02.07.2019 05:40 ybetancourt1

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Mathematics, 20.06.2019 18:04, vnfrancis3716
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!
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Mathematics, 21.06.2019 22:10, ansonferns983
Given: ae ≅ ce ; de ≅ be prove: abcd is a parallelogram. we have that ab || dc. by a similar argument used to prove that △aeb ≅ △ced, we can show that △ ≅ △ceb by. so, ∠cad ≅ ∠ by cpctc. therefore, ad || bc by the converse of the theorem. since both pair of opposite sides are parallel, quadrilateral abcd is a parallelogram.
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