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Circles
Given: ABCD is a quadrilateral whose diagonals AC and BD intersect each other at right angles at O.
To Prove: Quadrilateral ABCD is a rhombus.
Proof: In ∆AOB and ∆AOD,
AO = AO | Common
OB = OD | Given
∠AOB = ∠AOD | Each = 90°
∴ ∆AOB ≅ ∆AOD
| SSS Congruence Rule
∴ AB = AD ...(1) | C.P.C.T.
Similarly, we can prove that
AB = BC ...(2)
BC = CD ...(3)
CD = AD ...(4)
In view of (1), (2), (3) and (4), we obtain
AB = BC = CD = DA
∴ Quadrilateral ABCD is a rhombus.
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(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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