In the given figure two circles intersect each other in points A and B. Secants through the points A intersect the circles in point P,Q and R,S.Line PR and line SQ intersect in point T.
Show that
(i) PTQ anf PBQ are supplementary
(ii) BSTR is a cyclic quadrilateral

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Solution

$(i) In △ RTS, ∠ RTS + ∠ TRS + ∠ TSR = 180 ° . . . . . (1) (Angle sum property) Now, ∠ TRS + ∠ PRS = 180 ° . . . . . . . . . (2) (Linear pair) ∠ PRS + ∠ PBA = 180 ° . . . . . (3) (ARPB is a cyclic quadrilateral) Subtracting eq (3) from eq (2) we get, ∠ TRS + ∠ PRS - ∠ PRS - ∠ PBA = 0 \Rightarrow ∠ TRS = ∠ PBA . . . . . (4) ∠ ASQ = ∠ ABQ (Angles i n the same segment) Putting the value of ∠ TRS and ∠ ASQ in equation (1) we get, ∠ RTS + ∠ TRS + ∠ TSR = 180 ° \Rightarrow ∠ RTS + ∠ PBA + ∠ ABQ = 180 ° \Rightarrow ∠ RTS + ∠ PBQ = 180 ° or, ∠ PTQ + ∠ PBQ = 180 °$

$(ii) In △ PTQ, ∠ PTQ + ∠ TPQ + ∠ PQT = 180 ° . . . . . (1) (angle sum property) Now, ∠ PQT + ∠ AQS = 180 ° . . . . . . . . . (2) (Linear pair) ∠ AQS + ∠ ABS = 180 ° . . . . . (3) (AQSB is a cyclic quadrilateral) Subtracting eq (3) from eq (2) we get, ∠ PQT + ∠ AQS - ∠ AQS - ∠ ABS = 0 \Rightarrow ∠ PQT = ∠ ABS . . . . . (4) ∠ RPA = ∠ RBA (angles in the same segment) Putting the value of ∠ TQP and ∠ PQT in equation (1) we get, ∠ PTQ + ∠ RBA + ∠ ABS = 180 ° \Rightarrow ∠ PTQ + ∠ RBS = 180 ° o r ∠ RTQ + ∠ RBS = 180 °$

Since, the sum of opposite angles of a quadrilateral is supplementary.
Hence, BSTR is a cyclic quadrilateral.

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In the given figure two circles intersect each other in points A and B. Secants through the points A intersect the circles in point P,Q and R,S.Line PR and line SQ intersect in point T. Show that (i) PTQ anf PBQ are supplementary (ii) BSTR is a cyclic quadrilateral

In the given figure two circles intersect each other in points A and B. Secants through the points A intersect the circles in point P,Q and R,S.Line PR and line SQ intersect in point T.
Show that
(i) PTQ anf PBQ are supplementary
(ii) BSTR is a cyclic quadrilateral