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.]
Given ∆ABC, lines are drawn through A, B and C parallel respectively to the sides
BC, CA and AB forming ∆PQR. Show that BC =
To Prove: ∆ABC ≅ ∆CDA
Proof: BC || DA
| Opposite sides of a parallelogram are parallel
and AC is a transversal
∴ ∠BCA = ∠DAC ...(1)
| Pair of alternate interior angles
Also, AB || DC
Opposite sides of a parallelogram are parallel
and AC is a transversal
∴ ∠BAC = ∠DCA ...(2)
| Pair of alternate interior angles
AC = CA ...(3) | Common
In view of (1), (2) and (3),
∆ABC ≅ ∆CDA
| ASA congruence criterion
ABCD is a trapezium in which AB || CD and AD = BC. Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ABC ≅ ∆BAD.