Question

in fig AB||CD XY intersects these lines at p&q if PR &QS are bisector of angle apx and angle Cqx then prove that PR||QS​

Answer 1

Step-by-step explanation:

As per diagram AB∣∣CD and PQ is transversal

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]∠CML=∠ALP=100

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]∠CML=∠ALP=100 o

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]∠CML=∠ALP=100 o [ Corresponding angles]

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]∠CML=∠ALP=100 o [ Corresponding angles]∠ALP=∠MLB=100

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]∠CML=∠ALP=100 o [ Corresponding angles]∠ALP=∠MLB=100 o

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]∠CML=∠ALP=100 o [ Corresponding angles]∠ALP=∠MLB=100 o [ Vertically opposite angles]

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]∠CML=∠ALP=100 o [ Corresponding angles]∠ALP=∠MLB=100 o [ Vertically opposite angles]∠CMQ=180−100=80

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]∠CML=∠ALP=100 o [ Corresponding angles]∠ALP=∠MLB=100 o [ Vertically opposite angles]∠CMQ=180−100=80 o

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]∠CML=∠ALP=100 o [ Corresponding angles]∠ALP=∠MLB=100 o [ Vertically opposite angles]∠CMQ=180−100=80 o [sum of angle on a straight line is 180^o$$]

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]∠CML=∠ALP=100 o [ Corresponding angles]∠ALP=∠MLB=100 o [ Vertically opposite angles]∠CMQ=180−100=80 o [sum of angle on a straight line is 180^o$$]∠CMQ=∠LMD=80

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]∠CML=∠ALP=100 o [ Corresponding angles]∠ALP=∠MLB=100 o [ Vertically opposite angles]∠CMQ=180−100=80 o [sum of angle on a straight line is 180^o$$]∠CMQ=∠LMD=80 o

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]∠CML=∠ALP=100 o [ Corresponding angles]∠ALP=∠MLB=100 o [ Vertically opposite angles]∠CMQ=180−100=80 o [sum of angle on a straight line is 180^o$$]∠CMQ=∠LMD=80 o [ Vertically opposite angles]

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]∠CML=∠ALP=100 o [ Corresponding angles]∠ALP=∠MLB=100 o [ Vertically opposite angles]∠CMQ=180−100=80 o [sum of angle on a straight line is 180^o$$]∠CMQ=∠LMD=80 o [ Vertically opposite angles]∠LMD=∠PLB=80

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]∠CML=∠ALP=100 o [ Corresponding angles]∠ALP=∠MLB=100 o [ Vertically opposite angles]∠CMQ=180−100=80 o [sum of angle on a straight line is 180^o$$]∠CMQ=∠LMD=80 o [ Vertically opposite angles]∠LMD=∠PLB=80 o

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]∠CML=∠ALP=100 o [ Corresponding angles]∠ALP=∠MLB=100 o [ Vertically opposite angles]∠CMQ=180−100=80 o [sum of angle on a straight line is 180^o$$]∠CMQ=∠LMD=80 o [ Vertically opposite angles]∠LMD=∠PLB=80 o [ Corresponding angles]

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]∠CML=∠ALP=100 o [ Corresponding angles]∠ALP=∠MLB=100 o [ Vertically opposite angles]∠CMQ=180−100=80 o [sum of angle on a straight line is 180^o$$]∠CMQ=∠LMD=80 o [ Vertically opposite angles]∠LMD=∠PLB=80 o [ Corresponding angles]∠PLB=∠ALM=80

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]∠CML=∠ALP=100 o [ Corresponding angles]∠ALP=∠MLB=100 o [ Vertically opposite angles]∠CMQ=180−100=80 o [sum of angle on a straight line is 180^o$$]∠CMQ=∠LMD=80 o [ Vertically opposite angles]∠LMD=∠PLB=80 o [ Corresponding angles]∠PLB=∠ALM=80 o

As per diagram AB∣∣CD and PQ is transversal∠DMQ=100 o [ Given]∠DMQ=∠CML=100 o [ Vertically opposite angles]∠CML=∠ALP=100 o [ Corresponding angles]∠ALP=∠MLB=100 o [ Vertically opposite angles]∠CMQ=180−100=80 o [sum of angle on a straight line is 180^o$$]∠CMQ=∠LMD=80 o [ Vertically opposite angles]∠LMD=∠PLB=80 o [ Corresponding angles]∠PLB=∠ALM=80 o [ Vertically opposite angles]