Long Answer TypeIn ∆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
Short Answer TypeABCD 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.]
Long Answer TypeShort Answer TypeLong Answer TypeShort Answer TypeGiven ∆ABC, lines are drawn through A, B and C parallel respectively to the sides
BC, CA and AB forming ∆PQR. Show that BC = 
Long Answer TypeGiven: ABCD is a square. AC and BD are its diagonals.
To Prove: AC = BD; AC ⊥ BD
Proof: In ∆ABC and ∆BAD,
AB = BA | Common
∠ABC = ∠BAD | Each = 90°
BC = AD
| Sides of a square are equal
∴ ∆ABC ≅ ∆BAD
| SAS congruence criterion
∴ AC = BD | CPCT
Again, in ∆AOB and ∆AOD,
AO = AO | Common
AB = AD
| Sides of a square are equal
OB = OD
| A square is a parallelogram and the diagonals of a parallelogram bisect each other
∴ ∆AOB ≅ ∆AOD
| SSS congruence criterion
∴ ∠AOB = ∠AOD | CPCT
But ∠AOB + ∠AOD = 180°
| Linear Pair Axiom
∴ ∠AOB = ∠AOD = 90°
⇒ AO ⊥ BD
⇒ AC ⊥ BD.
ABCD is a trapezium in which AB || CD and AD = BC. Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ABC ≅ ∆BAD.
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