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Circles
Question
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ABCD is a trapezium in which AB || CD and AD = BC. Show that
(i)    ∠A = ∠B
(ii)    ∠C = ∠D
(iii)    ∆ABC ≅ ∆BAD.

Answer

Given: ABCD is a trapezium in which AB || CD and AD = BC

To Prove:
(i) ∠A = ∠B
(ii)    ∠C = ∠D
(iii)    ∆ABC ≅ ∆BAD

Construction: Through C, draw CE || DA to intersect AB produced at E.

Proof:
(i) ∵ AB || CD    | Given
and AD || CE    | by construction
∴ AECD is a parallelogram
| A quadrilateral is a parallelogram if both the pairs of opposite sides are parallel
∴ AD = EC
| Opposite sides of a parallelogram are equal
But AD = BC    | Given
∴ BC = EC
∴ ∠1 = ∠2
| Angles opposite to equal sides of a triangle are equal
∠B + ∠2 = 180°    ...(1)
| Linear Pair Axiom
∵ AD || EC    | by construction
and AE intersects them
∴ ∠A + ∠1 = 180°    ...(2)
| Sum of the consecutive interior angles on the same side of a transversal is 180°
From (1) and (2),
∠B + ∠2 = ∠A + ∠1
But ∠1 = ∠2    | Proved above
∴ ∠B = ∠A
⇒ ∠A = ∠B
(ii)    ∵ AB || DC    | Given
and AD is a transversal
∴∠A + ∠D = 180°    ...(3)
| Sum of the consecutive interior angles on the same side of a transversal is 180°
∵ AB || DC    | Given
and BC is a transversal.
∴ ∠B + ∠C = 180°    ...(4)
| Sum of the consecutive interior angles on the same side of a transversal is 180°
From (3) and (4),
∠A + ∠D = ∠B + ∠C
But ∠A = ∠B    | Proved in (i
∴ ∠D = ∠C
⇒ ∠C = ∠D.
(iii)    In ∆ABC and ∆BAD,
AB = BA    | Common
BC = AD    | Given
∠ABC = ∠BAD    | Proved in (i)
∴ ∆ABC ≅ ∆BAD
| SAS congruence criterion

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Some More Questions From Circles Chapter

Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that:
(i)    it bisects ∠C also
(ii)    ABCD is a rhombus.

ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that (i) ABCD is a square (ii) diagonal BD bisects ∠B as well as ∠D.

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that:

(i)    ∆APD ≅ ∆CQB
(ii)   AP = CQ
(iii)  ∆AQB ≅ ∆CPD
(iv)  AQ = CP
(v)   APCQ is a parallelogram.

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD respectively (see figure). Show that:
(i) ∆APB ≅ ∆CQD
(ii) AP = CQ.

In ∆ABC and ∆DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, Band C are joined to vertices D, E and F respectively (see figure). Show that:
(i)     quadrilateral ABED is a parallelogram
(ii)    quadrilateral BEFC is a parallelogram
(iii)   AD || CF and AD = CF
(iv)   quadrilateral ACFD is a parallelogram



(v)     AC = DF
(vi)    ∆ABC ≅ ∆DEF. [CBSE 2012

ABCD is a trapezium in which AB || CD and AD = BC (see figure): Show that
(i)      ∠A = ∠B
(ii)    ∠C = ∠D
(iii)    ∆ABC = ∆BAD
(iv)    diagonal AC = diagonal BD.



[Hint. Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles.