Xét ΔEAD và ΔEBC có
\(\widehat{EAD}=\widehat{EBC}\)(hai góc so le trong, AD//BC)
\(\widehat{AED}=\widehat{BEC}\)(hai góc đồng vị)
Do đó:ΔEAD~ΔEBC
=>\(\dfrac{EA}{EB}=\dfrac{AD}{BC}\)
=>\(\dfrac{2.2}{x}=\dfrac{3}{5}\)
=>\(x=\dfrac{2,2\cdot5}{3}=\dfrac{11}{3}\)