
Let us draw BM \(⊥\) PQ and CN \(⊥\) RS.
As PQ \(||\) RS,
Therefore, BM \(||\) CN
Thus, BM and CN are two parallel lines and a transversal line BC cut them at B and C respectively.
\(∴ \) \(∠\)2 = \(∠\)3 (Alternate interior angles)
However, \(∠\)1 = \(∠\)2 and \(∠\)3 = \(∠\)4 (By laws of reflection)
∴ \(∠\)1 = \(∠\)2 = \(∠\)3 = \(∠\)4
Also, \(∠\)1 + \(∠\)2 = \(∠\)3 + \(∠\)4
\(∠\)ABC = \(∠\)DCB
However, these are alternate interior angles.
∴ AB \(|| \)CD




Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 9.27). Prove that ∠ACP = ∠ QCD

ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.14). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ABC ≅ ∠∆BAD
(iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

(i) The kind of person the doctor is (money, possessions)
(ii) The kind of person he wants to be (appearance, ambition)