Hello, abc123!

1) $\displaystyle ABCD{E}F$ is a regular hexagon and $\displaystyle EDGH$ is a square.

Find the measures of: $\displaystyle \angle EFH,\;\angle FHD,\;\angle DFH$ Code:

A * - - - * F
/ :\*
/ : \ *
/ : \ *
B * 120° : E * *
\ : /90° * *
\ : / * H
\ :/ /
C * - - - * /
D * /
*
G

All interior angles of the hexagon are 120°.

We have: .$\displaystyle \angle FED = 120^o,\;\angle DEH = 90^o$

. . Hence: .$\displaystyle \angle FEH = 150^o$

$\displaystyle \Delta EFH$ is isosceles, $\displaystyle EF = EH$

. . Therefore: .$\displaystyle \angle EFH \:=\:\angle EHF \:=\:15^o$

In square $\displaystyle EDGH,\;\angle EHD = 45^o$

. . Therefore: .$\displaystyle \angle FHD \:=\:15^o + 45^o \:=\:60^o$

$\displaystyle \angle AFE = 120^o,\;\angle AFD = 90^o\quad\Rightarrow\quad \angle DFE = 30^o$

. . Therefore: .$\displaystyle \angle DFH \:=\:30^o + 15^o \:=\:45^o$