Question

in the diagram below, m∠cih = 100° and m∠bgd = 43°. find m∠ahf.
step angle reason
1 m∠cih = 100° given
2 m∠bgd = 43° given
3 m∠blank = blank° sum of angles in a quadrilateral
you may assume lines that appear straight are straight, but the figure is not otherwise drawn to scale

Explanation:

Step1: Identify vertical - angles

$\angle BGD$ and $\angle AGF$ are vertical - angles. So, $m\angle AGF=m\angle BGD = 43^{\circ}$ (Vertical angles are equal).

Step2: Identify linear - pair

$\angle CIH$ and $\angle HIG$ form a linear - pair. So, $m\angle HIG = 180^{\circ}-m\angle CIH=180 - 100=80^{\circ}$ (Linear - pair of angles are supplementary).

Step3: Use angle - sum property of a triangle

In $\triangle FHG$, let's consider the fact that the sum of angles in a triangle is $180^{\circ}$. In the quadrilateral formed by the intersection of the lines, we can also use the angle - sum property of a quadrilateral. But an alternative way is to note that $\angle AHF$ and $\angle GHF$ are supplementary. First, in the triangle formed by the intersection of the lines with vertices at $H$, $G$, and the intersection point of the lines containing $AF$ and $CE$ (let's call it a virtual triangle), we know that $\angle AHF$ and $\angle AGF+\angle HIG$ are related.
We know that the sum of angles around a point is $360^{\circ}$. Consider the intersection point $H$. Let's use the fact that $\angle AHF + \angle AGF+\angle HIG=180^{\circ}$ (by the property of angles formed by intersecting lines).
Substitute the values of $\angle AGF = 43^{\circ}$ and $\angle HIG = 80^{\circ}$ into the equation:
$m\angle AHF=180-(43 + 80)$
$m\angle AHF = 57^{\circ}$

Answer:

$57^{\circ}$