12. ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD
Solution: To Construct: Draw a line through C parallel to DA intersecting AB produced at E. (i) CE = AD (Opposite sides of a parallelogram) AD = BC (Given) , BC = CE ⇒∠CBE = ∠CEB also, ∠A+∠CBE = 180° (Angles on the same side of transversal and ∠CBE = ∠CEB) ∠B +∠CBE = 180° ( As Linear pair) ⇒∠A = ∠B (ii) ∠A+∠D = ∠B+∠C = 180° (Angles on the same side of transversal) ⇒∠A+∠D = ∠A+∠C (∠A = ∠B) ⇒∠D = ∠C (iii) In ΔABC and ΔBAD, AB = AB (Common) ∠DBA = ∠CBA AD = BC (Given) , ΔABC ≅ ΔBAD [SAS congruency] (iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBAD.
