, 2015

Solution:

Since $D$ and $E$ are the midpoints of the segments $BC$ and $AC$ the lines $DE$ and $AB$ are parallel. It follows that $\angle EDA = \angle BAD$, and by the Angles Subtended by Same Arc Theorem we have $\angle BAD = \angle BAP = \angle BQP$. Therefore

$\angle EQP = \angle EDA = \pi - \angle PDE$ which means that the quadrilateral $EDPQ$ is cyclic, and the triangles $TED$ and $TPQ$ are similar. Let $T$ be the centroid of the triangle $ABC$ and $x = |DP| = |EQ|$. From the similarity of the triangles $TED$ and $TPQ$ we deduce
$$\frac{|TD| + x}{|TE|} = \frac{|TP|}{|TE|} = \frac{|TQ|}{|TD|} = \frac{|TE| + x}{|TD|}$$
We rearrange this to
$$(|TD| - |TE|)(|TD| + |TE| + x) = 0,$$
which gives $|TE| = |TD|$ since $|TD| + |TE| + x > 0$. The triangle $DET$ is therefore isosceles with apex at vertex $T$. Since the lines $AD$ and $BE$ are parallel the triangle $ABT$ is also isosceles with apex at vertex $T$, hence $|BE| = |BT| + |TE| = |AT| + |TD| = |AD|$. From this it follows that the triangles $ABE$ and $BAD$ are congruent since they have two pairs of sides of the same length and an angle of the same size between them $\angle EBA = \angle BAD$. Thus $|AE| = |BD|$ and hence $|AC| = |BC|$.

As in the first solution we prove that the lines $DE$ and $AB$ are parallel and that the quadrilateral $EDPQ$ is cyclic. Therefore $\angle EPD = \angle EQD$. Since the chords $DP$ and $EQ$ are of equal length we have $\angle DEP = \angle QDE$. Hence $\angle PDE = \angle DEQ$ which means that the quadrilateral $EDPQ$ is an isosceles trapezoid. Thus $\angle TED = \angle EDT = \angle BAT$ and so $\angle BED = \angle BAD$. This means that the quadrilateral $ABDE$ is cyclic. Since the lines $AB$ and $ED$ are parallel the quadrilateral $ABDE$ is also isosceles trapezoid. It follows $|AE| = |BD|$ and hence $|AC| = |BC|$.