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13/07/2021 12:11 pm
Topic starter
ABCD is a trapezium in which AB || CD and AD = BC (see Figure). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ΔABC ≅ ΔBAD
(iv) diagonal AC = diagonal BD
1 Answer
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13/07/2021 12:13 pm
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
∠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 ≅ ΔBA.