Question

Consider the rhombus ABCD with the side length AB of 14 cm and the measure of the angle B of 120 °. Through the vertex A goes a certain secant that intersects the extensions of the sides BC and CD in E, respectively F.
Calculate:
$\frac{1}{ce} + \frac{1}{cf}$
​

Answer 1

Answer:

  1/14

Step-by-step explanation:

In the given geometry, we have similar triangles:

  ΔCEF ~ ΔBEA ~ ΔDAF

This lets us write the relations for corresponding sides ...

  CF/CE = DF/DA

where DF = CF -CD

We can rearrange this relation to give the value we're looking for as follows.

__

Making the substutition for DF, and multiplying by CE, we have ...

  CF = CE(CF -CD)/DA

Both DA and CD are sides of the rhombus, so are both 14 cm in length. Multiplying by DA and making the number substitution, we have ...

  DA·CF = CE·CF -CE·CD

  14·CF = CE·CF -14·CE . . . . substitute 14 for DA and CD

  14(CF +CE) = CE·CF . . . . . add 14·CE

  (CF +CE)/(CE·CF) = 1/14 . . . . divide by 14·CE·CF

  1/CE +1/CF = 1/14 . . . . . . . . . simplify