Two chords PQ and QR are equildistant form `O` the centre of the circle. If QS, is the diameter, then show that QS bisects `angle PQR` and `anglePSR`.

Explanation

Draw OM and ON perpendicular from `O` to PQ and QR respectively.
In `Delta QOM` and `DeltaQON`,
`because" "{(OM,=,,ON,("given")),(angleQMO,=,,angleQNO,("each"90^@)),(QO,=,,QO,("common")):}`
`therefore Delta QOM congDelta QON` (R.H.S.)
`rArrangle MQO=angle NQO`
Now, QS is the diameter of the circle. `thereforeangle QPS=angle QRS=90^@` (`becauseangle` in a semicircle is right angle)
In `DeltaQPS` and `DeltaQRS`,
`because" "{(angleQPS,=,,angleQRS=90^@,),(QS,=,,QS,("common")),(anglePQS,=,,angleRQS,("proved above")):}`
`thereforeDeltaQPScongDeltaQRS`
`rArrangle PSQ=angle RSQ`
`therefore QS`, bisects `angle PQR` and `angle PSR`.