ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig). Show that F is the mid-point of BC.

Let EF intersect DB at G.

By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side, bisects the third side.
In ∆ABD,
EF || AB and E is the mid-point of AD.
Therefore, G will be the mid-point of DB.
As EF || AB and AB || CD,
∠EF || CD (Two lines parallel to the same line are parallel to each other)
In ∆BCD, GF || CD and G is the mid-point of line BD.
Therefore, by using converse of mid-point theorem, F is the mid-point of BC.

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)