Final Round of National Olympiad

Figure 15

By assumptions, $\frac{|PA|}{|PB|}=\frac{|PD|}{|PC|}$. As the bases $AB$ and $CD$ are parallel, we have also $\frac{|PA|}{|PB|}=\frac{|PC|}{|PD|}$ (Fig. 15). Hence $|PC|=|PD|$. Similarity of triangles $APD$ and $BPC$ implies $\frac{|AD|}{|BC|}=\frac{|PD|}{|PC|}=1$, thus $|AD|=|BC|$ as needed.

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Alternative solution.

Figure 16

Figure 17

By assumptions, triangles $APD$ and $BPC$ are similar. Thus $\angle ADB=\angle ACB$ (Fig. 16), showing that quadrilateral $ABCD$ is cyclic. But if a quadrilateral with parallel opposite sides has a circumcircle, the bisectors of these sides coincide as they have the same direction and both pass through the circumcenter of the quadrilateral (Fig. 17). By symmetry w.r.t. this line, the other pair of opposite sides have equal lengths.