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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

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Short Answer Type

12.

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.]

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

2967 Views

Long Answer Type

14. AB and CD are two parallel lines and a transversal I intersects AB at X and CD at Y. Prove that the bisectors of the interior angles form a rectangle.

2190 Views

Short Answer Type

15. ABCD is a parallelogram and line segments AX, CY bisect the angles A and C respectively. Show that AX || CY.

3477 Views

Long Answer Type

16. If a diagonal of a parallelogram bisects one of the angles of the parallelogram, it also bisects the second angle and then the two diagonals are perpendicular to each other.

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Short Answer Type

17.

Given ∆ABC, lines are drawn through A, B and C parallel respectively to the sides

BC, CA and AB forming ∆PQR. Show that BC = $1 half.$

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18. “A diagonal of a parallelogram divides it into two congruent triangles.” Prove it.

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Long Answer Type

19. Show that the diagonals of a square are equal and perpendicular to each other.

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20.
ABCD is a trapezium in which AB || CD and AD = BC. Show that
(i)    ∠A = ∠B
(ii)    ∠C = ∠D
(iii)    ∆ABC ≅ ∆BAD.

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

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

Given: ABCD is a trapezium in which AB || CD and AD = BC
To Prove:(i)

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