Given -
- [tex]\sf{\angle{EFG}=105\degree}[/tex]
To prove-
Solution-
Assumption
AD || EH
If AD || EH and CG is a transversal.
Then,
[tex]\implies\sf{\angle{BFE+EFG}=180\degree\:(linear\:pair)}[/tex]
[tex]\implies\sf{\angle{BFE+105\degree}=180\degree}[/tex]
[tex]\implies\sf{\angle{BFE}=180\degree-\:105\degree}[/tex]
[tex]\implies\bf{\angle{BFE}=75\degree}[/tex]
We know,
[tex]\sf{\angle{BFE}=\angle{CBA}=75\degree\:(corresponding\:angles)}[/tex]
[tex]\sf{\angle{BFE}=x\degree=75\degree}[/tex]
[tex]\therefore\boxed{\bf{x\degree=75\degree}}[/tex]
