Question

in the diagram below, $overline{fg}congoverline{hg}$, $overline{bd}paralleloverline{ei}$ and $mangle acb = 61^{circ}$. find $mangle fgh$. you may assume lines that appear straight are straight, but the figure is not otherwise drawn to scale. answer attempt 2 out of 3 $mangle fgh=square^{circ}$

Explanation:

Step1: Identify vertical - angles

$\angle ACB$ and $\angle FCD$ are vertical - angles. So, $m\angle FCD=m\angle ACB = 61^{\circ}$.

Step2: Use parallel - lines property

Since $\overline{BD}\parallel\overline{EI}$, $\angle FCD$ and $\angle FGI$ are corresponding angles. Then $m\angle FGI=m\angle FCD = 61^{\circ}$.

Step3: Consider isosceles triangle

In $\triangle FGH$, $\overline{FG}\cong\overline{HG}$, so $\angle H=\angle GFH$. Let $m\angle H = m\angle GFH=x$.
In $\triangle FGH$, we know that the sum of interior angles of a triangle is $180^{\circ}$. Also, $\angle FGI$ is an exterior angle of $\triangle FGH$.
By the exterior - angle theorem of a triangle, $\angle FGI=\angle H+\angle GFH$.
Since $\angle H=\angle GFH$, and $\angle FGI = 61^{\circ}$, then $m\angle FGH=180^{\circ}- 2\times61^{\circ}$.
$m\angle FGH=180 - 122=58^{\circ}$.

Answer:

$58$