Find the measures of the following angles

Explanation

Strategy

To answer part a and b, we will use the property of vertical angles.  To answer part c, we will write an equation involving m(∠ABF) that mathematically models the situation.  To answer part d, use the property of vertical angles after you answer part c

WHY

For part a, we note that \(\overleftrightarrow{AD}\) and \(\overleftrightarrow{BC}\) intersect to form vertical angles.  For part b, note that \(\overleftrightarrow{FE}\) and \(\overleftrightarrow{BC}\) intersect to form vertical angles as well.  For part c, we can solve the equation to find the unknown m(∠ABF); then for part d, m(∠DBE) is simply equals to m(∠ABF) because they are vertical angles.

Solution

1. If we ignore \(\overleftrightarrow{FE}\) for the moment, we see that \(\overleftrightarrow{AD}\) and \(\overleftrightarrow{BC}\) intersect to form the pair of vertical angles ∠CBD and ∠1.  By the property of vertical angles,

   ∠CBD ≅ ∠1 (Read as "angle CBD is congruent to angle one.")

    Since congruent angles have the same measures,

   m(∠CBD) = m(∠1) 

   In the figure, we are given m(∠CBD) = 50^o.  Thus, m(∠1) can write m(∠1) = 50^o.

2. We see that (\overleftrightarrow{FE}\) and (\overleftrightarrow{BC}\) intersect to form the pair of vertical angles ∠FBC and ∠2.  By the property of vertical angles 

   ∠FBC ≅ ∠2  which means they have the same measure

   m(∠FBC) = m(2)

3. Since ∠ABC is a straignt angle, the sum of the measures of ∠ABF, the 100^o angle, and the 50^o angle is 180^o. If we let x = m(∠ABF), we have
   • x + 100^o + 50^o = 180^o
     • The word sum indicates addition
   • x + 150^o = 180^o
     • On the left side, combine like terms: 100^o + 50^o = 150^o
   • x = 30^o
     • To isolate x, undo the addition of 150^o by subtracting 150^o from both sides: 180^o - 150^o = 30^o

   Thus, m(∠ABF) = 30^o

4. I hope you noticed by now that ∠ABF and ∠DBE are vertical angles, which means their measures are equal.  From point c, you know that m(∠ABF) = 30^o

   m(∠DBE) = 30^o.