In the given fig. from an external point P, a tangent PT and a line segment PAB is drawn to a circle with centre O. ON is perpendicular on the chord AB. Prove (i) PA.PB = PN² – AN² (ii) PN² – AN² = OP² – PT² (iii) PA.PB = PT².
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SOLUTION :
Given:
PT is a tangent drawn from an external point P and a line segment PAB is drawn to a circle with Centre O.ON is perpendicular on the chord AB.
ON ⟂ AB
TO PROVE:
(i) PA.PB = PN² – AN²
(ii) PN² – AN² = OP² – PT²
(iii) PA.PB = PT².
CONSTRUCTION: In the given figure join OA.
PROOF IS IN THE ATTACHMENT
HOPE THIS WILL HELP YOU...
Given:
PT is a tangent drawn from an external point P and a line segment PAB is drawn to a circle with Centre O.ON is perpendicular on the chord AB.
ON ⟂ AB
TO PROVE:
(i) PA.PB = PN² – AN²
(ii) PN² – AN² = OP² – PT²
(iii) PA.PB = PT².
CONSTRUCTION: In the given figure join OA.
PROOF IS IN THE ATTACHMENT
HOPE THIS WILL HELP YOU...
Attachments:
Answered by
8
Step-by-step explanation:
Solution : (i) PA . PB = (PN – AN) (PN + BN) = (PN – AN) (PN + AN) (As AN = BN) = PN2 – AN2 (ii) PN2 – AN2 = (OP2 – ON2) – AN2 (As ON⊥PN) = OP2 – (ON2 + AN2) = OP2 – OA2 (As ON⊥AN) = OP2 – OT2 (As OA = OT) (iii) From (i) and (ii) PA.PB = OP2 – OT2 = PT2 (As 90 and bisrctor .hence proved
Similar questions
1. PA = (PN-AN) , PB = (PN+BN )
PA.PB = (PN-AN)(PN+BN)
ON Perpendicular AB so AN = BN
PA.PB = (PN-AN)(PN+AN) = PN^2-AN^2
2. PN^2-AN^2 =
IN ∆ONA
OA^2 = ON^2+AN^2
AN^2 = OA^2-ON^2
PN^2-(OA^2-ON^2)= PN^2+ON^2-OA^2
In ∆ ONP
OP^2 = ON^2+PN^2
so. OP^2- OA^2
OA= OT
PN^2-AN^2= OP^2-OT^2
3. from 1 & 2
PA.PB = OP^2-OT^2
IN ∆ OTP
OP^2 = PT^2+OT^2
OP^2-OT^2= PT^2
So PA.PB = PT^2