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  # ABCD is a trapezium in which AB || CD and AD = BC (see the given figure). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]
  
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  ## Let us extend AB. Then, draw a line through C, which is parallel to AD, intersecting AE at point E. It is clear that AECD is a parallelogram. (i) AD = CE (Opposite sides of parallelogram AECD) However, AD = BC (Given) Therefore, BC = CE ∠CEB = ∠CBE (Angle opposite to equal sides are also equal) Consider parallel lines AD and CE. AE is the transversal line for them. ∠A + ∠CEB = 180º (Angles on the same side of transversal) ∠A + ∠CBE = 180º (Using the relation∠CEB = ∠CBE) ... (1) However, ∠B + ∠CBE = 180º (Linear pair angles) ... (2) From equations (1) and (2), we obtain ∠A = ∠B (ii) AB || CD ∠A + ∠D = 180º (Angles on the same side of the transversal) Also, ∠C + ∠B = 180° (Angles on the same side of the transversal) ∴ ∠A + ∠D = ∠C + ∠B However, ∠A = ∠B [Using the result obtained in (i)] ∴ ∠C = ∠D (iii) In ΔABC and ΔBAD, AB = BA (Common side) BC = AD (Given) ∠B = ∠A (Proved before) ∴ ΔABC ≅ ΔBAD (SAS congruence rule) (iv) We had observed that, ΔABC ≅ ΔBAD ∴ AC = BD (By CPCT)
  
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