ABCD is a trapezium in which AB || CD & AD=BC. Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ΔABC ≅ ΔBAD
(iv) AC = BD
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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)