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Circles
Given: 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.
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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
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