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2016-02-27T17:55:33+05:30

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In quadrilateral APIQ we have,
∠A = 60° [ Given ]
∠CQA = 90° [ altitude of a triangle is a line segment
                       through a vertex and perpendicular ]
∠BPA = 90° [ altitude of a triangle is a line segment                                    through a vertex and perpendicular ]

⇒ ∠A + ∠CQA + ∠BPA + ∠QIP = 360° [ ASP of a quad. ]
⇒ 60° + 90° + 90° + ∠QIP = 360° 
⇒ ∠QIP = 120°
and, ∠QIP + ∠A = 120° + 60°
    ⇒  ∠QIP + ∠A = 180°
And we know that,
 If a pair of opp. angles of a quad. is supplementary then the quad. is cyclic.
∴ AQIP is a cyclic quadrilateral.      HENCE PROVED.

In quadrilateral PBQC we have,
∠BQC = 90° [ altitude of a triangle is a line segment
                       through a vertex and perpendicular ]

∠CPB = 90° [ altitude of a triangle is a line segment
                       through a vertex and perpendicular ]

which implies to theorem,
Angle in the same segment of a circle are equal.
that means, PBQC is a cyclic quadrilateral.  
  HENCE PROVED.
3 5 3
you can view the diagram of this que. right below and when you'll
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but it is given that it is not perpendicular