Similar Triangles

Okay. Let's work this through step by step. If $m\angle {BAC} = 70^\circ$, and $\triangle{ABC} \sim \triangle {PAQ}$, then $\angle{APQ}$ should also be $70^\circ$.

If $\angle{APQ} = 70^\circ , \angle{CPQ} = 110^\circ$.

And since $\triangle{QPC} \sim \triangle{ABQ} , \angle{AQB} = 110^\circ$.

Also, since $\angle{AQB} = 110^\circ , \angle{AQC} = 70^\circ$.

Okay, now for the second part.

Since $\triangle {ABQ} \sim \triangle {QPC} , \angle {PQC} = \angle {BAQ}$.

Let both these angles be $x$.

This means that $\angle {QAP} = \angle {AQP}$, since $\angle {BAC} = 70^\circ$ and $\angle{AQC} = 70^\circ$, and both $\angle {QAP}$  and $\angle{AQP}$ equal $(70-x)^\circ$.

Since $\angle {APQ} = 70^\circ$, both  $\angle {QAP}$  and $\angle{AQP}$ are equal to $55^\circ$.

Now the final step. Now that we have the angle measurements of $\angle {AQB}$ and $\angle {AQP}$, we can solve for $\angle {PQC}$.

$\angle {AQB} + \angle {AQP} = 165^\circ$.

 $\angle {PQC} = 180^\circ - 165^\circ = 15^\circ$

$\boxed {m\angle {PQC} = 15^\circ}$

Since triangle ABC  is similar to triangle PAQ

Then angle BAC  = angle APQ

But angle BAC  = 70

So angle APQ  = 70

And angle QPC  is supplemental to angle APQ

So angle QPC  = 110

And triangle ABQ  is simialr to triangle QCP

So  angle ABQ = angle QCP  

But angle ABQ  = angle ABC = angle QCP = angle ACB

So...triangle ABC  isosceles  with angles ABC, ACB  being equal

So...their measures  =  [ 180  - m angle BAC ] / 2  =   [180  - 70] / 2  = 110 / 2  = 55

And since m angle ACB  = m angle QCP...then m angle QCP  = 55

So...in triangle QPC......m angle PQC  = [ 180  - m angle QPC - m angle QCP ] =

180  - 110  - 55  = 180 - 165  =   15°  =  m angle PQC

Thank you so much guys.