In the given figure,$AB \perp BQ$,$PQ \perp QB$,$AC = PR$ and $BR = QC$. Prove that $\angle BAC = \angle QPR$.

(N/A) Given: $AB \perp BQ$,$PQ \perp QB$,$AC = PR$ and $BR = QC$.
Since $AB \perp BQ$,$\angle ABQ = 90^{\circ}$. Thus,$\angle ABC = 90^{\circ}$.
Since $PQ \perp QB$,$\angle PQB = 90^{\circ}$. Thus,$\angle PQR = 90^{\circ}$.
Given $BR = QC$.
Adding $RC$ to both sides,we get $BR + RC = QC + RC$.
Therefore,$BC = QR$......$(1)$
In $\Delta ABC$ and $\Delta PQR$:
$AC = PR$ (Given)
$\angle ABC = \angle PQR = 90^{\circ}$ (Proved above)
$BC = QR$ [From $(1)$]
By $RHS$ congruence criterion,$\Delta ABC \cong \Delta PQR$.
Therefore,$\angle BAC = \angle QPR$ (by $CPCT$).