In the figure, \(∠Q > ∠R\), \(PB\) is the angle-bisecting line segment of \(∠QPR\) and \(PN ⊥ QR\). Show that \(∠BPN=\frac{1}{2}(∠Q−∠R)\).
Proof:
\(PB\) is angle-bisecting line segment of \(∠QPR\) .
Therefore, \(∠QPB=∠BPR\) ----- (1)
In \(△ PQN\), by angle sum relation in a triangle,
\(∠PQN+∠PNQ+∠QPN=\)\(^°\)
\(∠PQN+\)\(^°+∠QPN=\)\(^°\)
\(∠PQN=\)\(^°−∠QPN\)
\(∠Q=\)\(^°−∠QPN\) --------(2)
Again, in \(△ PNR\), by angle sum relation in a triangle,
\(∠PNR+∠PRN+∠RPN=\)\(^°\)
\(^°+∠PRN+∠RPN=\)\(^°\)
\(∠PRN=\)\(^°−∠RPN\)
\(∠R=\)\(^°−∠RPN\) -------.(3)
Subtracting (3) from (2), and simplifying then we get,
\(=2∠BPN\) (From (1))
\(∠BPN=\frac{1}{2} [∠Q−∠R]\)