Question

I drew two circles which intersect each other at the points G and H. Now I drew a straight
line through the point G which intersects 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 Rand S. Let
us prove that PQ = RS.​

Answer 1

Refer pic.

Refer pic. Join AC,PQ and BD.

Refer pic. Join AC,PQ and BD.ACQP is a cyclic quadrilaterl.

Refer pic. Join AC,PQ and BD.ACQP is a cyclic quadrilaterl.∠CAP+∠PQC=180

Refer pic. Join AC,PQ and BD.ACQP is a cyclic quadrilaterl.∠CAP+∠PQC=180 ∘

Refer pic. Join AC,PQ and BD.ACQP is a cyclic quadrilaterl.∠CAP+∠PQC=180 ∘ ....(1) pair of opposites in cyclic quadrilateral.

Refer pic. Join AC,PQ and BD.ACQP is a cyclic quadrilaterl.∠CAP+∠PQC=180 ∘ ....(1) pair of opposites in cyclic quadrilateral.PQDB is a cyclic quadrilaterl.

Refer pic. Join AC,PQ and BD.ACQP is a cyclic quadrilaterl.∠CAP+∠PQC=180 ∘ ....(1) pair of opposites in cyclic quadrilateral.PQDB is a cyclic quadrilaterl.∠PQD+∠DBP=180

Refer pic. Join AC,PQ and BD.ACQP is a cyclic quadrilaterl.∠CAP+∠PQC=180 ∘ ....(1) pair of opposites in cyclic quadrilateral.PQDB is a cyclic quadrilaterl.∠PQD+∠DBP=180 ∘

Refer pic. Join AC,PQ and BD.ACQP is a cyclic quadrilaterl.∠CAP+∠PQC=180 ∘ ....(1) pair of opposites in cyclic quadrilateral.PQDB is a cyclic quadrilaterl.∠PQD+∠DBP=180 ∘ ....(2) pair of opposites in cyclic quadrilateral.

Refer pic. Join AC,PQ and BD.ACQP is a cyclic quadrilaterl.∠CAP+∠PQC=180 ∘ ....(1) pair of opposites in cyclic quadrilateral.PQDB is a cyclic quadrilaterl.∠PQD+∠DBP=180 ∘ ....(2) pair of opposites in cyclic quadrilateral.∠PQC+∠PQD=180

Refer pic. Join AC,PQ and BD.ACQP is a cyclic quadrilaterl.∠CAP+∠PQC=180 ∘ ....(1) pair of opposites in cyclic quadrilateral.PQDB is a cyclic quadrilaterl.∠PQD+∠DBP=180 ∘ ....(2) pair of opposites in cyclic quadrilateral.∠PQC+∠PQD=180 ∘

Refer pic. Join AC,PQ and BD.ACQP is a cyclic quadrilaterl.∠CAP+∠PQC=180 ∘ ....(1) pair of opposites in cyclic quadrilateral.PQDB is a cyclic quadrilaterl.∠PQD+∠DBP=180 ∘ ....(2) pair of opposites in cyclic quadrilateral.∠PQC+∠PQD=180 ∘ .....(3) CQD is a straight line.

Refer pic. Join AC,PQ and BD.ACQP is a cyclic quadrilaterl.∠CAP+∠PQC=180 ∘ ....(1) pair of opposites in cyclic quadrilateral.PQDB is a cyclic quadrilaterl.∠PQD+∠DBP=180 ∘ ....(2) pair of opposites in cyclic quadrilateral.∠PQC+∠PQD=180 ∘ .....(3) CQD is a straight line.From (1), (2) and (3), we get,

Refer pic. Join AC,PQ and BD.ACQP is a cyclic quadrilaterl.∠CAP+∠PQC=180 ∘ ....(1) pair of opposites in cyclic quadrilateral.PQDB is a cyclic quadrilaterl.∠PQD+∠DBP=180 ∘ ....(2) pair of opposites in cyclic quadrilateral.∠PQC+∠PQD=180 ∘ .....(3) CQD is a straight line.From (1), (2) and (3), we get,∠CAP+∠DBP=180

Refer pic. Join AC,PQ and BD.ACQP is a cyclic quadrilaterl.∠CAP+∠PQC=180 ∘ ....(1) pair of opposites in cyclic quadrilateral.PQDB is a cyclic quadrilaterl.∠PQD+∠DBP=180 ∘ ....(2) pair of opposites in cyclic quadrilateral.∠PQC+∠PQD=180 ∘ .....(3) CQD is a straight line.From (1), (2) and (3), we get,∠CAP+∠DBP=180 ∠CAB+∠DBA=180

Refer pic. Join AC,PQ and BD.ACQP is a cyclic quadrilaterl.∠CAP+∠PQC=180 ∘ ....(1) pair of opposites in cyclic quadrilateral.PQDB is a cyclic quadrilaterl.∠PQD+∠DBP=180 ∘ ....(2) pair of opposites in cyclic quadrilateral.∠PQC+∠PQD=180 ∘ .....(3) CQD is a straight line.From (1), (2) and (3), we get,∠CAP+∠DBP=180 ∠CAB+∠DBA=180 If a traversal intersects 2 lines such that a pair of interior angles on same side of traversal is supplimentary, then the 2 lines are parallel.

Refer pic. Join AC,PQ and BD.ACQP is a cyclic quadrilaterl.∠CAP+∠PQC=180 ∘ ....(1) pair of opposites in cyclic quadrilateral.PQDB is a cyclic quadrilaterl.∠PQD+∠DBP=180 ∘ ....(2) pair of opposites in cyclic quadrilateral.∠PQC+∠PQD=180 ∘ .....(3) CQD is a straight line.From (1), (2) and (3), we get,∠CAP+∠DBP=180 ∠CAB+∠DBA=180 If a traversal intersects 2 lines such that a pair of interior angles on same side of traversal is supplimentary, then the 2 lines are parallel.∴AC∥BD