Question

In Fig. 14.71, APB and COD are straight lines
through the points of intersection of two circles.
Prove (i) AC || BD, (ii) ZCPD = ZAQB. (SC)​

Answer 1

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Join AC, PQ and BD

ACQP is a cyclic quadrilateral

∴ ∠CAP + ∠PQC = 180° ...(i)

(pair of opposite in a cyclic quadrilateral are supplementary)

PQDB is a cyclic quadrilateral

∴ ∠PQD + ∠DBP = 180° ….(ii)

(pair of opposite angles in a cyclic quadrilateral are supplementary)

Again, ∠PQC + ∠PQD = 180° ... (iii)

(CQD is a straight line

Using (i), (ii) and (iii)

∴ ∠CAP + ∠DBP = 180° Or

∴ ∠CAB + ∠DBA =180°

We know, if a transversal intersects two lines such That a pair of interior angles on the same side of the Transversal is supplementary, then the two lines are parallel

∴ AC || BD

Answer 2

Join AC, PQ and BD

ACQP is a cyclic quadrilateral

∴ ∠CAP + ∠PQC = 180° ...(i)

(pair of opposite in a cyclic quadrilateral are supplementary)

PQDB is a cyclic quadrilateral

∴ ∠PQD + ∠DBP = 180° ….(ii)

(pair of opposite angles in a cyclic quadrilateral are supplementary)

Again, ∠PQC + ∠PQD = 180° ... (iii)

(CQD is a straight line

Using (i), (ii) and (iii)

∴ ∠CAP + ∠DBP = 180° Or

∴ ∠CAB + ∠DBA =180°

We know, if a transversal intersects two lines such That a pair of interior angles on the same side of the Transversal is supplementary, then the two lines are parallel

∴ AC || BD

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