I saw your pencil writings on the question. Those workings are correct and will lead to our answers eventually.

Realise that when P and B are midpoints of AQ and AC respectively, triangles APB and AQC would be similar. This is because

- the ratio AP/AQ = the ratio AB/AC = 1/2 (obviously, since P and B are midpoints)
- angle PAB = angle QAC (the same angle, of course!)

We have uses the Side-Angle-Side (SAS) proof of similarity for this. This theorem is important and have earned itself the name "midpoint theorem".

We can easily see that angles APB and AQC would be equal by virtue of this similarity.

This would mean that angle PQC must be "x".

Since angle QPC is "180° - x", we can see that angle PCQ
= 180° - angle QPC - angle PQC
= 180° - (180° - 2x) - x
= x

Then it's clear that angles PQC and PCQ are equal and therefore triangle PCQ is isosceles.