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Given: ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.
To Prove: (i) ABCD is a square.
(ii) diagonal BD bisects ∠B as well as ∠D.
Proof: (i) ∵ AB || DC
and transversal AC intersects them.
∴ ∠ACD = ∠CAB | Alt. Int. ∠s
But ∠CAB = ∠CAD
∴ ∠ACD = ∠CAD
∴ AD = CD
| Sides opposite to equal angles of a triangle are equal
∴ ABCD is a square.
(ii) In ∆BDA and ∆DBC,
BD = DB | Common
DA= BC
| Sides of a square ABCD
AB = DC
| Sides of a square ABCD
∴ ∆BDA ≅ ∆DBC
| SSS Congruence Rule
∴ ∠ABD = ∠CDB | C.P.C.T.
But ∠CDB = ∠CBD
| ∵ CB = CD (Sides of a square ABCD)
∴ ∠ABD = ∠CBD
∴ BD bisects ∠B.
Now, ∠ABD = ∠CBD
∠ABD = ∠ADB | ∵ AB = AD
∠CBD = ∠CDB | ∵ CB = CD
∴ ∠ADB = ∠CDB
∴ BD bisects ∠D.
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Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that:
(i) it bisects ∠C also
(ii) ABCD is a rhombus.
(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.]
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