Question

Segment AB is a diameter of a circle with Centre P segment AC is a chord a secent through p and parallel to segment AC intersect the tangent drawn at C in D prove that line DB is a tangent to the circle

Answer 1

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Step-by-step explanation:

Answer 2

Given : Segment AB is a diameter of a circle with Centre P segment AC is a chord a secent through p

To prove :  DB is the tangent

Explanation:

in triangle ACP

PA = CP  (radii)

∠ACP = ∠CAP

(angle opp to equal side are equal )

AC║ PD

∠CAP = ∠DPB (corresponding angles)

∠ACP = ∠CPD (alternate interior angle)

∠DPB = ∠CPD

now , in triangle

DCP and DBP

PC = PB (radii)

PD = PD (common)

∠DPB = ∠CPD (proved above )

thus

$\bigtriangleup DCP \cong \bigtriangleup DBP$

by (SAS congruency )

therefore ,

∠PCD = ∠PBD (by CPCT)

now,

∠PCD = 90° (radius is perpendicular to the tangent)

∠PCD = ∠PBD = 90°

thus , DB is the tangent

hence proved

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