MHB Finding $\dfrac{AC}{BD}$ of a Trapezoid $ABCD$

[Albert1]

A trapezoid $ABCD$ , $BC//AD ,\,\, \angle A=90^o$ , $AC\perp BD$

given :$\dfrac {BC}{AD}=k$

find :$ \dfrac {AC}{BD}$

 

[mente oscura]

A trapezoid $ABCD$ , $BC//AD ,\,\, \angle A=90^o$ , $AC\perp BD$

given :$\dfrac {BC}{AD}=k$

find :$ \dfrac {AC}{BD}$

Hello.

Spoiler

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.

 

[Albert1]

Hello.

Spoiler

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.

very good :)