# find AC/BD

#### Albert

##### Well-known member
A trapezoid $ABCD$ , $BC//AD ,\,\, \angle A=90^o$ , $AC\perp BD$

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

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

#### mente oscura

##### Well-known member
A trapezoid $ABCD$ , $BC//AD ,\,\, \angle A=90^o$ , $AC\perp BD$

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

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

#### Albert

##### Well-known member
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.
very good