Two circles intersect each other at the points G and H . A straight line is drawn through the point G which intersect two circles at the points P and Q and the straight line through the point H parallel to PQ intersects the two circles at the points R and S . Prove that PQ=RS .

Explanation

Let us join P, R and Q, S
As per question , PQ||RS and PS is their transversal.
`thereforeangleSPQ=angleRSP" …..(1) "[because"alternate angles"] `
Similarly , PQ ||RS and GH is their transversal.
`thereforeanglePGH=angleSHG` ……..(2) [`because` alternate angles]
Again, PRHG is a cyclic quadrilateral
`thereforeanglePGH+anglePRH=180^(@)` .........(3)
Similarly , QSHG is a cyclic quadrilateral,
`angleSQG+angleSHG=180^(@)` ............(4)
From (3) and (4) we get , `anglePGH+anglePRH=angleSQG+angleSHG`
or, `anglePRH=angleSQG` [From(2)]
or,` anglePRS=angleSQP` ................(5)
Now , in `Delta PQS,anglePSR=angleQPS` [from(1)] `anglePRS=angleSQP` [From (5) ] and PS is common to both
`thereforeDeltaPRScongDeltaPQS` [by the A-A-S condition of conguency]
`thereforeRS=PQ [because` similar sides of congruent triangles]
Hence PQ=RS