Question

In the diagram below, AB I AC, CD= CE and m B = 48º.Find m DEB.

Answer 1

In the given triangle the sum of the interior angles of triangle ACB is equal to 180 degrees, therefore, we have the following relationship:

$\angle CAB+\angle ABC+\angle BCA=180$

Replacing the values we get:

$90+48+\angle BCA=180$

Solving the operations:

$138+\angle BCA=180$

Now we solve for angle BCA by substracting 138 to both sides:

$\begin{gathered} \angle BCA=180-138 \\ \angle BCA=42 \end{gathered}$

Now, since segments CD and CE are equal, this means that triangle CDE is an isosceles triangle, therefore, its base angles are the same, and we have the following relationship:

$\angle CDE+\angle DEC+\angle ECD=180$

Replacing the known values:

$2\angle DEC+42=180$

Now we solve for angle DEC by subtracting 42 to both sides:

$\begin{gathered} 2\angle DEC=180-42 \\ 2\angle DEC=138 \end{gathered}$

Now we divide both sides by 2:

$\begin{gathered} \angle DEC=\frac{138}{2} \\ \angle DEC=69 \end{gathered}$

Now, angles DEC and DEB are supplementary, therefore, their sum adds up to 180, therefore, we have:

$\angle DEC+\angle\text{DEB}=180$

Replacing the known angle:

$69+\angle DEB=180$

Subtracting 69 to both sides:

$\begin{gathered} \angle DEB=180-69\rbrack \\ \angle DEB=111 \end{gathered}$

Therefore, angle DEB measures 111 degrees.