11th Junior Balkan Mathematical Olympiad

In the rays $DA$ and $BA$ we take a point $E$ and $Z$, respectively, such that $AC=AE=AZ$.

Since $\widehat{DEC}=\frac{\widehat{DAC}}{2}=18^{\circ }=\widehat{CBD}$, the quadrilateral $DEBC$ is cyclic.

Similarly, the quadrilateral $CBZD$ is cyclic, because $$\widehat{AZC}=\frac{\widehat{BAC}}{2}=36^{\circ }=\widehat{BDC}.$$

Figure 8 Figure 9

Therefore the pentagon $BCDZE$ is inscribed in the circle $K(A,AC)$. It gives $AC=AD$ and $\widehat{ACD}=\widehat{ADC}=\frac{180^{\circ }-36^{\circ }}{2}=72^{\circ }$, which gives $\widehat{ADP}=36^{\circ }$ and $\widehat{APD}=108^{\circ }$.

Alternative solution: See the figure in the right. If $C_1$ is the symmetric point of $D$ with respect to the line $BC$, then $AB{C}_{1}D$ is inscribable. Since $\widehat{BA{C}_1}=72^{\circ }$ and $\widehat{BD{D}_1}=36^{\circ }$ we have that $D{D}_1$ is bisector of the angle $BD{C}_1$ and hence $C_1$ is the incenter of the triangle $BD{C}_1$. Therefore we have $$\widehat{D{C}_{1}A}=\widehat{A{C}_{1}B}=36^{\circ }\text{and}\widehat{DPA}=\frac{72^{\circ }+144^{\circ }}{2}=108^{\circ }.$$