From a point $P$ outside a circle, draw two tangents to the circle touching at points $A$ and $B$. Draw a sectant line intersecting the circle at points $C$ and $D$, with $C$ between $P$ and $D$. Choose point $Q$ on the chord $CD$ such that $\angle DAQ=\angle PBC$. Prove that $\angle DBQ=\angle PAC$.
My Attempt
I have somehow made a figure but could not solve it. Please help me with this..
Let O be the center of the circle ACBD.
Fact-1: All the same color coded angles are equal.
Fact-2: The red dotted circle will pass through A,P, B, O.
Fact-3: $\angle purple1 = \angle green + \angle red$, a standard result from tangent properties.
By considering $\triangle ADQ$, $\angle blue = \angle green + \angle red$.
Since $\angle purple1 = \angle blue$, we say that A, P, B, Q, O are con-cyclic.
Then, $\angle black = \angle purple2 = \angle purple1$
That means $\angle green + \angle yellow = \angle green + \angle red$.
Result follows.