in the figure ABCD is a trapezium in which AB|| CD and AD=BC show that angle a=angle b
Answers
(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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Given :- ABCD is a trapezium
AB || CD
AD = BC
To proof :-
(i)∠A = ∠B
(ii)∠C = ∠D
(iii)∆ ABC ≅ ∆ BAD
(iv)Diagonal AC = Diagonal = BD
Construction :- Draw DA || CE
Solution :-
(i) Since it's given ABCD is a trapezium
AB || CD
DA || CE ( By construction)
Therefore, ADCE is a parallelogram
So, DA = CE &
DC = AE ( Opposite side of parallelogram are equal )
But, AD = BC
Therefore, BC = CE ( Given )
∠CEB = ∠CBE ( In ∆ CBE angles opposite to equal sides are equal )
180° - ∠DAB = 180° - ∠ABC
[ ADCE is a parallelogram and ∠A + ∠E = 180° ∠B & ∠CBE form a linear pair ]
∠A = ∠B ( Cancelling 180° from both sides)
(ii) Co interior angles on the same side of a transversal are supplementary
∠A + ∠D = 180° & ∠B + ∠C = 180°
∠A + ∠D = ∠B + ∠C
∠B + ∠D = ∠B + ∠C ( ∠A = ∠B proved above)
∠D = ∠C
(iii) In ∆ ABC & ∆ BAD
AB = BA
∠B = ∠A ( proved above )
BC = BD ( Given )
∆ ABC ≅ ∆ BAD ( By SAS criteria)
(iv) AC = BD ( CPCT )
