Question

In trapezul ABCD ,AB paralel CD,AB mai mic CD ,se considera E apartine AD si F apart BC,astfel incat AE supra ED ⇒3 PE 4 si EF paralel AB.stiind ca BC 14 cm,calculati FC si BF
rezolvarea problemei cu Teorema lui Thales
va rog e urgent!!!

Answer 1

Prin teorema paralelelor neechidistante rezulta ca:
$\frac{AE}{ED}= \frac{BF}{FC}$⇒$\frac{BF}{FC}= \frac{3}{4}$⇒
$\frac{BF+FC}{FC}= \frac{3+4}{4}$⇒$\frac{14}{FC}= \frac{7}{4}$⇒$FC= \frac{14*4}{7}=8cm$
BF=BC-FC
BF=14-8
BF=6cm