$\frac{AC}{BD} = \frac{AB.AD+BC.CD}{AB.BC+CD.AD} \Leftrightarrow \frac{AC.sinA}{BC.sinB}=\frac{AB.AD.sinA+BC.CD.sinD}{AB.BC.sinB+CD.AD.sinC}=\frac{2S_{ABCD}}{2S_{ABCD}}=1 \Leftrightarrow \frac{AC}{BD}=\frac{sinB}{sinA}=\frac{\frac{sinB}{sinACB}}{\frac{sinA}{sinADB}}=\frac{\frac{AC}{AB}}{\frac{BD}{AB}}$
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