Step-by-step explanation: To draw a line perpendicular to AB through A and B, respectively. Use the following steps of construction. 1.Draw a line segment AB = 4 cm. 2.Taking 4 as centre and radius more than ½ AB (i.e., 2 cm) draw an arc say it intersect AB at E. 3.Taking E as centre and with same radius as above draw an arc which intersect previous arc at F. Again, taking F as centre and with same radius as above draw an arc which intersect previous arc (obtained in step ii) at G. 5.Taking G and F are centres, draw arcs which intersect each other at H. 6.Join AH . Thus, AX is perpendicular to AB at A. Similarly, draw BY ⊥ AB at B. Now, we know that if two lines are parallel, then the angle between them will be 0° or 180°. Here, ∠XAB = 90° [∴ XA ⊥ AB] and ∠YBA = 90° [ ∴ YB ⊥ AB] ∠XAB+ ∠YBA = 90° + 90° = 180° So, the lines XA and YS are parallel. [since, it sum of interior angle on same side of transversal is 180°, then the two lines are parallel] Reply

Step-by-step explanation:To draw a line perpendicular to AB through A and B, respectively. Use the following steps of construction.

1.Draw a line segment AB = 4 cm.

2.Taking 4 as centre and radius more than ½ AB (i.e., 2 cm) draw an arc say it intersect AB at E.

3.Taking E as centre and with same radius as above draw an arc which intersect previous arc at F.

Again, taking F as centre and with same radius as above draw an arc which intersect previous arc (obtained in step ii) at G.

5.Taking G and F are centres, draw arcs which intersect each other at H.

6.Join AH . Thus, AX is perpendicular to AB at A. Similarly, draw BY ⊥ AB at B.

Now, we know that if two lines are parallel, then the angle between them will be 0° or

180°.

Here, ∠XAB = 90° [∴ XA ⊥ AB]

and ∠YBA = 90° [ ∴ YB ⊥ AB]

∠XAB+ ∠YBA = 90° + 90° = 180°

So, the lines XA and YS are parallel.

[since, it sum of interior angle on same side of transversal is 180°, then the two lines are

parallel]