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 TypeGiven: ABCD is a ||gm. Diagonal AC bisects ∠A.
To Prove: (i) AC bisects ∠C
(ii) AC ⊥ BD.
Proof: (i) ∵ AB || DC and AC intersects them
∴ ∠1 = ∠4
| Alternate Interior ∠s
Similarly, ∠2 = ∠3
| Alternate Interior ∠s
But ∠1 = ∠2
∴ ∠3 = ∠4
⇒ AC bisects ∠C.
(ii) In ∆ADC,
∠2 = ∠4
∴ AD = CD
| Sides opposite to equal angles of a triangle are equal
In ∆AOD and ∆COD,
OA = OC | ∵ Diagonals
of a ||gm bisect each other
OD = OD | Common side
AD = CD | Proved above
∴ ∆AOD ≅ ∆COD
| SSS Congruence Axiom
∴ ∠AOD = ∠COD | C.P.C.T.
But ∠AOD + ∠COD = 180°
| Linear Pair Axiom
⇒ 2∠AOD = 180°
| ∵ ∠AOD = ∠COD
⇒ ∠AOD = 90°
⇒ AC ⊥ BD.
Short 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 TypeABCD 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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