Question

In the given figure $DE||BC,AB||EF$ and $\angle ABC={60}^o$. Find the $\angle ADF,\angle EFD$ and $\angle FDE$.

Answer 1

REF.Image

Solution : To find : $\angle ADF,\angle EFD,\angle FDE$

By mid-point theorem :

$AD=DB=EF$

$DE\left|\right|BC,DB$ is a traversal,

$\angle ADE=\angle ABC={60}^{\circ }$ [corresponding $\angle s$]

$\angle FED={60}^{\circ }$

$BFED$ is a parallelogram :

$\therefore \angle BDE=\angle BFE={120}^{\circ }$

$\therefore \angle ADF={120}^{\circ }$

Since $△EFD$ is an equilateral $△$

from mid-point theorem

$\therefore \angle EFD=\angle FDE={60}^{\circ }$