Exercise 30 Page 330

It is given that $P$ is the incenter of the triangle $△AEC.$ By definition, an incenter is the point of intersection of the three angle bisectors of a triangle. Hence, $\overline{AD},$ $\overline{CF},$ and $\overline{EB}$ are the angle bisectors of $△AEC.$ First, let's consider $\overline{AD}.$ It is a bisector of $\angle EAC,$ so angles $\angle EAD$ and $\angle DAC$ are congruent. Thus, the measure of $\angle DAB$ is also $33^{\circ }.$ Now, we can find the measure of $\angle EAC.$ By the Angle Addition Postulate, its measure equals the sum of $m\angle EAD$ and $m\angle DAC.$ Now we are going to consider $\overline{CF}.$ It is a bisector of the angle $\angle ACE,$ so $\angle ACF$ and $\angle FCE$ are congruent. Therefore, $\angle ACF$ also measures $28.5^{\circ }.$ We can find the measure of $\angle ACE$ by adding the measures of $\angle ACF$ and $\angle FCE.$ Now that we know two of the angles measures in $△AEC,$ $\angle EAC$ and $\angle ACE,$ we can use the Angle Sum Theorem. This theorem tells us that the angle measures in any triangle add up to $180^{\circ }.$ Now we will consider the last bisector, $\overline{EB}.$ It bisects the angle $\angle AEC,$ so $\angle AEB$ and $\angle CEB$ are congruent angles. This means that their measures are the same. We can use this to write an equation that relates the bisected angles to the larger angle. Also, we earlier found that $\angle AEC$ measures $57^{\circ }.$ Let's solve it! Therefore, the measure of $\angle AEB,$ as well as $\angle CEB,$ is $28.5^{\circ }.$ Note that $\angle CEB$ and $\angle DEP$ are both names for the same angle. Hence, the answer is $28.5^{\circ }.$