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
In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.