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find AC/BD

Albert

Well-known member
Jan 25, 2013
1,225
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
Nov 29, 2013
172
A trapezoid $ABCD$ , $BC//AD ,\,\, \angle A=90^o$ , $AC\perp BD$

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

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

[tex]If \ \angle{A}=90º \ and \ \overline{BC} // \overline{AD} \rightarrow{}\angle{B}=90º[/tex]

[tex]If \ \overline{BC} // \overline{AD} \rightarrow{}\angle{ADB}=\angle{DBC}= \alpha[/tex]

[tex]\sin{\alpha}=\dfrac{\overline{AB}}{\overline{BD}}=\dfrac{\overline{BC}}{\overline{AC}}[/tex]

[tex]\cos{\alpha}=\dfrac{\overline{AB}}{\overline{AC}}=\dfrac{\overline{AD}}{\overline{BD}}[/tex]

Therefore:

[tex]\overline{AB}=\dfrac{\overline{BC} \ \overline{BD}}{\overline{AC}}[/tex]

[tex]\overline{AB}=\dfrac{\overline{AD} \ \overline{AC}}{\overline{BD}}[/tex]

[tex]\dfrac{\overline{BC} \ \overline{BD}}{\overline{AC}}=\dfrac{\overline{AD} \ \overline{AC}}{\overline{BD}}[/tex]


[tex]\dfrac{\overline{BC}}{\overline{AD}}= \dfrac{(\overline{AC})^2}{(\overline{BD})^2}=k[/tex]

Therefore:

[tex]\dfrac{\overline{AC}}{\overline{BD}}=\sqrt{k}[/tex]


Regards.
 

Albert

Well-known member
Jan 25, 2013
1,225
Hello.

[tex]If \ \angle{A}=90º \ and \ \overline{BC} // \overline{AD} \rightarrow{}\angle{B}=90º[/tex]

[tex]If \ \overline{BC} // \overline{AD} \rightarrow{}\angle{ADB}=\angle{DBC}= \alpha[/tex]

[tex]\sin{\alpha}=\dfrac{\overline{AB}}{\overline{BD}}=\dfrac{\overline{BC}}{\overline{AC}}[/tex]

[tex]\cos{\alpha}=\dfrac{\overline{AB}}{\overline{AC}}=\dfrac{\overline{AD}}{\overline{BD}}[/tex]

Therefore:

[tex]\overline{AB}=\dfrac{\overline{BC} \ \overline{BD}}{\overline{AC}}[/tex]

[tex]\overline{AB}=\dfrac{\overline{AD} \ \overline{AC}}{\overline{BD}}[/tex]

[tex]\dfrac{\overline{BC} \ \overline{BD}}{\overline{AC}}=\dfrac{\overline{AD} \ \overline{AC}}{\overline{BD}}[/tex]


[tex]\dfrac{\overline{BC}}{\overline{AD}}= \dfrac{(\overline{AC})^2}{(\overline{BD})^2}=k[/tex]

Therefore:

[tex]\dfrac{\overline{AC}}{\overline{BD}}=\sqrt{k}[/tex]


Regards.
very good :)