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Circles
Given: ABCD is a rhombus. P, Q, R, S are the mid-points of AB, BC, CD, DA respectively. PQ, QR, RS and SP are joined.
To Prove: PQRS is a rectangle.
Construction: Join AC and BD.
Proof: In triangles RDS and PBQ,
DS = QB
| Halves of opposite sides of || gm ABCD which are equal
DR = PB
| Halves of opposite sides of || gm ABCD which are equal
∠SDR = ∠QBP
| Opposite ∠s of ||gm ABCD which are equal
∴ ∆RDS = ∆PBQ
| SAS Congruence Axiom
∴ SR = PQ | C.P.C.T.
In triangles RCQ and PAS,
RC = AP
| Halves of opposite sides of || gm ABCD which are equal
CQ = AS
Halves of opposite sides of || gm ABCD which are equal
∠RCQ = ∠PAS
Opposite ∠s of || gm ABCD which are equal
∴ ∆RCQ ≅ ∆PAS
| SAS Congruence Axiom
∴ RQ = SP | C.P.C.T.
∴ In PQRS,
SR = PQ and RQ = SP
∴ PQRS is a parallelogram,
In ∆CDB,
∵ R and Q are the mid-points of DC and CB respectively.
∴ RQ || DB ⇒ RF || EO.
Similarly, RE || FO
∴ OFRE is a || gm
∴ ∠R = ∠EOF = 90°
| ∵ Opposite ∠s of a || gm are equal and diagonals of a rhombus intersect at 90°
Thus PQRS is a rectangle.
Sponsor Area
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.]
Sponsor Area
Sponsor Area