Balti Tee 1998 · Ülesanne 13

Solution:

Figure 4

Let $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ be the circumcircles of triangles $A E D$ and $B C D$, respectively. Let $D P$ meet $\mathcal{C}_{2}$ for the second time at $F$ (see Figure 4). Since $\angle A D E=\angle B D C$, the ratio of the lengths of the segments $E A$ and $B C$ is equal to the ratio of the radii of $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$. Thus the homothety with centre $P$ that takes $A E$ to $C B$, also transforms $\mathcal{C}_{1}$ onto $\mathcal{C}_{2}$. The same homothety transforms the arc $D E$ of $\mathcal{C}_{1}$ onto the $\operatorname{arc} F B$ of $\mathcal{C}_{2}$. Therefore $\angle E A D=\angle B D F=\angle B D P$. The second equality is proved similarly.