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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 cyclic quadrilaterals

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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.

∠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.