, 2012

Since $AB = BC$, $CD = DE$ we have $\sim AB = \sim BC$, $\sim CD = \sim DE$.

Therefore,
$$\angle BPA = \angle DPE = \frac{1}{2}(\sim AB + \sim DE) = \frac{1}{2}(\sim AB + \sim CD) = \angle BQA. \quad (1)$$

So, points $A$, $B$, $Q$, $P$ belong to the same circle and we have
$$\angle QAB = \angle QPB \text{ (subtended by the same chord $BQ$)} \quad (2)$$
and
$$\angle QAB = \angle CAB = \angle CEB \text{ (subtended by the same chord $BC$).} \quad (3)$$
Thus from (2) and (3) we have $\angle QPB = \angle CEB$, so $PQ \parallel CE$. Therefore,
$$\angle TQP = \angle DTE = \frac{1}{2}(\sim BC + \sim DE) = \frac{1}{2}(\sim AB + \sim CD) = \angle AQB. \quad (4)$$
Similarly, points $P$, $T$, $D$, $E$ belong to the same circle because
$$\angle EPD = \frac{1}{2}(\sim AB + \sim DE) = \frac{1}{2}(\sim BC + \sim DE) = \angle ETD.$$
So, $\angle PTE = \angle PDE$ (subtended by the same chord $PE$). Since $\angle ACE = \angle ADE$ (subtended by the same chord $AE$) we have $\angle ACE = \angle PTE$. Therefore, $PT \parallel AC$ and hence $\angle PTQ = \angle AQB$. Now from (4) it follows that $\angle PTQ = \angle TQP$, thus triangle $PTQ$ is isosceles.