The discussion focuses on calculating the ratio $\dfrac{AC}{BD}$ in trapezoid $ABCD$, where $BC$ is parallel to $AD$, $\angle A$ is $90^\circ$, and $AC$ is perpendicular to $BD$. Given the ratio $\dfrac{BC}{AD} = k$, the objective is to derive the expression for $\dfrac{AC}{BD}$. The geometric properties of trapezoids and the relationships between the sides are crucial for this calculation.
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Albert said:A trapezoid $ABCD$ , $BC//AD ,\,\, \angle A=90^o$ , $AC\perp BD$
given :$\dfrac {BC}{AD}=k$
find :$ \dfrac {AC}{BD}$
very good :)mente oscura said:Hello.
If \ \angle{A}=90º \ and \ \overline{BC} // \overline{AD} \rightarrow{}\angle{B}=90º
If \ \overline{BC} // \overline{AD} \rightarrow{}\angle{ADB}=\angle{DBC}= \alpha
\sin{\alpha}=\dfrac{\overline{AB}}{\overline{BD}}=\dfrac{\overline{BC}}{\overline{AC}}
\cos{\alpha}=\dfrac{\overline{AB}}{\overline{AC}}=\dfrac{\overline{AD}}{\overline{BD}}
Therefore:
\overline{AB}=\dfrac{\overline{BC} \ \overline{BD}}{\overline{AC}}
\overline{AB}=\dfrac{\overline{AD} \ \overline{AC}}{\overline{BD}}
\dfrac{\overline{BC} \ \overline{BD}}{\overline{AC}}=\dfrac{\overline{AD} \ \overline{AC}}{\overline{BD}}\dfrac{\overline{BC}}{\overline{AD}}= \dfrac{(\overline{AC})^2}{(\overline{BD})^2}=k
Therefore:
\dfrac{\overline{AC}}{\overline{BD}}=\sqrt{k}
Regards.