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In a triangle ABC if angle A=60degree and altitudes from b and c mmet acAC and AB at p and q and intersect each other at I prove that APIQ and PBQC are c

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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.