Show that ABQC is a cyclic quadrilateral & $\triangle BPD $ is isosceles

Question

Hi can anyone help me out here , Because I struggling in iii & iv

What I have done so far

$i.\angle BAC=2a$

$ii. \angle BAQ= \angle QAC$

$\therefore \angle BCQ = \angle QAC$ (alternate segment theorem)

& $ \angle BCQ = \angle QAC $

Answer 1

For (iii) :

All we need to prove is that $\angle{BAQ}=\angle{BCQ}$. Use that $\triangle{BQC}$ is an isosceles triangle.

For (iv) :

We have that $\angle{BDP}=\angle{BAP}$ and that $\angle{DBP}=\angle{DAP}$ from which $\angle{BDP}=\angle{DBP}$ follows.