In the given figure \(\angle QPR = 90^{\circ}\), \(PS\) is its bisector. If \(SX \perp PR\), establish that \(SX \times (PQ + PR) = PQ \times PR\).
Proof:
In \(\triangle PQR\), \(PS\) is the bisector of \(\angle P\).
By angle bisector theorem, we have:
Adding \(1\) on both sides, we get:
---- (\(1\))
Consider \(\triangle RXS\) and \(\triangle RPQ\).
\(\angle RXS =\)
\(\angle SRX =\)
By \(AA\) Similarity criterion, we have:
Then,
---- (\(2\))
Using (\(2\)) in (\(1\)), we have:
\(SX \times (PQ + PR) = PQ \times PR\)
Hence, we proved.
Answer variants:
\(\frac{PQ}{PR} = \frac{QS}{SR}\)
\(\frac{PR + PQ}{PR} = \frac{SR + QS}{SR}\)
\(\angle QRP = \angle R\)
\(1 + \frac{PQ}{PR} = 1 + \frac{QS}{SR}\)
\(\frac{PR + PQ}{PR} = \frac{QR}{SR}\)
\(\angle RPQ = 90^{\circ}\)