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Circles
Given: ABCD is a square.
To Prove: (i) AC = BD
(ii) AC and BD bisect each other at right angles.
Proof: (i) In ∆ABC and ∆BAD,
AB = BA | Common
BC = AD Opp. sides of square ABCD
∠ABC = ∠BAD | Each = 90°
(∵ ABCD is a square)
∴ ∆ABC ≅ ∆BAD
| SAS Congruence Rule
∴ AC = BD | C.P.C.T
(ii) In ∆OAD and ∆OCB,
AD = CB
| Opp. sides of square ABCD
∠OAD = ∠OCB
| ∵ AD || BC and transversal AC intersects them
∠ODA = ∠OBC
| ∵ AD || BC and transversal BD intersects them
∴ ∆OAD ≅ ∆OCB
| ASA Congruence Rule
∴ OA = OC ...(1)
Similarly, we can prove that
OB = OD ...(2)
In view of (1) and (2),
AC and BD bisect each other.
Again, in ∆OBA and ∆ODA,
OB = OD | From (2) above
BA = DA
| Opp. sides of square ABCD
OA = OA | Common
∴ ∆OBA ≅ ∆ODA
| SSS Congruence Rule
∴ ∠AOB = ∠AOD | C.P.C.T.
But ∠AOB + ∠AOD = 180°
| Linear Pair Axiom
∴ ∠AOB = ∠AOD = 90°
∴ AC and BD bisect each other at right angles.
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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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