P is any point on diameter of a circle with centre O.A perpendicular drawn on diameter at the point O intersects the point O intersects the circle at the point Q. Extended QP intersects the circle at the point R.A tangent drawn at the point R intersects extended OP at the point S. Prove theat `SP=SR. [GP-X]`

Explanation

P is any point on a diameter EF of the circle with centre O. `OQ bot EF and OQ` intersects the circel at the point Q. Extended QP intersects the circle at the point R. The tangent TS of the circle at R intersects extended OP at the point S.
To prove : `SP =SR.`
Construction : Let us join O, R.
Proof : OR is a radius of the circle passing through the point of
contact of the tangent `SRT. therefore OR bot ST.`
`therefore angle ORS =90^(@)or angle OEQ+angle QRS=90^(@)........(1)`
Again `OQ bot EStherefore angle QOS=90^(@), therefore angle OQP+angle OPQ=90^(@)..(2)`
from (1) and (2) we get, `angle ORQ+ angle QRS+angle OQP..(3)`
Now `OQ=OR[because` radii of same circle] `thereforeangle ORQ= angle OQR`
`therefore` from (3) we get, `angle OQR+angle QRS=angle OQP +angleOPQ[because angle ORQ =angle OQR]`
or , `angle OQP +angle QRS =angle OQR+ angle SPR [because angle OPQ ="opposite"angle SPR]`
or `angle QRS=angle SPRor, angle PRS =angle SPR`
`therefore SP = SR[because` in `DeltaSPR,` sides opposite of equal angle are equal]
`therefore SP =SR` (Proved)