10.3. Inside the isosceles triangle $ABC (AC = BC)$ with $m(\angle C) = 80^{\circ}$, there is a point $P$ such that $m(\angle PAB) = 30^{\circ}$ and $m(\angle PBA) = 10^{\circ}$. Determine the measure in degrees of the angle $CPB$.
Solution: In $\square A B C, m(\angle A)=m(\angle B)=\frac{180^{\circ}-80^{\circ}}{2}=50^{\circ}$, so $m(\angle P B D)=50^{\circ}-10^{\circ}=40^{\circ}$.

$\angle B P D$ - exterior $\square A B P \Rightarrow m(\angle B P D)=30^{\circ}+10^{\circ}=40^{\circ}$, therefore $\square P B D$ - isosceles $\Rightarrow P D=B D$.

$\angle P D C$ - exterior $\square P D B \Rightarrow m(\angle P D C)=40^{\circ}+40^{\circ}=80^{\circ}$, therefore $\square A C D$ - isosceles $\Rightarrow A C=A D$.

From $A D=B C$ and $P D=B D$ it follows that $A P=C D$.

We draw the altitude $C M$ in $\square A B C \Rightarrow$

$\square A F B$ - isosceles. $\angle P F A$ - exterior $\square A F B$

$\Rightarrow m(\angle P F A)=10^{\circ}+10^{\circ}=20^{\circ}$, and

![](https://cdn.mathpix.com/cropped/2024_06_06_20168aaaf2f9e4db9e97g-2.jpg?height=317&width=792&top_left_y=595&top_left_x=1054)
$m(\angle P A F)=30^{\circ}-10^{\circ}=20^{\circ}$, so $\square A P F$ - isosceles $\Rightarrow A P=P F \Rightarrow P F=C D$. $\angle E P F$ - exterior $\square A P F \Rightarrow m(\angle E P F)=20^{\circ}+20^{\circ}=40^{\circ}, m(\angle E C D)=80^{\circ}: 2=40^{\circ}, m(\angle F E P)=m(\angle D E C)$ $\Rightarrow m(\angle P F E)=m(\angle C D E)$. Therefore, by the SAS criterion $\square C E D \Rightarrow \square P E F \Rightarrow P E=C E$ $\Rightarrow \square P C E$ - isosceles. $\angle P E C$ - exterior $\square C E D \Rightarrow m(\angle P E C)=40^{\circ}+80^{\circ}=120^{\circ}$ $\Rightarrow m(\angle C P E)=\left(180^{\circ}-120^{\circ}\right): 2=30^{\circ} . m(\angle C P B)=m(\angle C P E)+m(\angle E P F)=30^{\circ}+40^{\circ}=70^{\circ}$.

Answer: $m(\angle C P B)=70^{\circ}$.