Question

În trapezul ABCD cu AB paralel cu CD , AB=18cm ,CD=6cm, AC=12cm . o punctul de intersectie al diagonalelor. atunci lungimea lui AO = ?

Answer 1

Ducem OE||AB||CD, E∈BC
In ΔBCD avem :
$\frac{OE}{CD} = \frac{BE}{BC}$
$\frac{BE}{BC}= \frac{BC-CE}{BC} = \frac{BC}{BC}- \frac{CE}{BC} = 1-\frac{CE}{BC}$
=>$\frac{OE}{CD} =1- \frac{CE}{BC}$
dar, In ΔCAB avem: 
$\frac{CE}{BC} = \frac{OE}{AB} =\frac{OE}{18}$
=> $\frac{OE}{CD}=\frac{OE}{6} =1- \frac{OE}{18}$
=>  $\frac{OE}{6} + \frac{OE}{18} =1=\ \textgreater \ \frac{3OE}{18} + \frac{OE}{18} =1=\ \textgreater \ \frac{4OE}{18} =1$
=> OE=4,5

$\frac{CO}{CA} = \frac{CO}{12} = \frac{OE}{AB} = \frac{OE}{18} = \frac{4,5}{18}$ =>
CO= 3
=> AO=CA-CO= 12-3=9
=> AO=9