
It is given that AB\( || \)CD and CD \(||\) EF
∴ AB\(||\) CD \(||\)EF (Lines parallel to the same line are parallel to each other)
It can be observed that x = z (Alternate interior angles) … (1)
It is given that y: z = 3: 7
Let the common ratio between y and z be a.
y = 3a and z = 7a
Also, x + y = 180º (Co-interior angles on the same side of the transversal)
z + y = 180º [Using equation (1)]
7a + 3a = 180º
10a = 180º
a = 18º
∴ x = 7a = 7 × 18º = 126º




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)