Question

plz help me with the following questions ;
prove that
1) PA×PB=PN^2- AN^2
2)PN^2-AN^2=OP^2-OT^2
3)PA×PB=PT^2​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

$\:\:\:\: \: \: \: \: \large\mathfrak{\underline{\huge\mathcal{\bf{\boxed{\boxed{\huge\mathcal{~~QUESTION~~}}}}}}}$

Prove=

1) PA×PB=PN²- AN²

2)PN²-AN²=OP²-OT²

3)PA×PB=PT²

$\:\:\:\: \: \: \: \: \large\mathfrak{\underline{\huge\mathcal{\bf{\boxed{\boxed{\huge\mathcal{~~AnsweR~~}}}}}}}$

_____________________________________________

$<font color=red><marquee behavior=alternate>Figure</marquee></font>$

is in the attachment...

$<font color=red><marquee behavior=alternate>given</marquee></font>$

•from an external point p ,a tangent and a line PAB is drawn to the circle of centre o.

$<font color=red><marquee behavior=alternate>To prove</marquee></font>$

1) PA×PB=PN²- AN²

2)PN²-AN²=OP²-OT²

3)PA×PB=PT²

$<font color=red><marquee behavior=alternate>proof</marquee></font>$

•

PA=PN-AN

PB=PN+BN

now...AN=BN..(cause ON_|_AB, so ON bisects AB)

now.

PA×PB=(PN-AN)(PN+BN)

=>PA×PB=(PN-AN)(PN+AN)

=>PA×PB=PN²-AN² (proved)

•

PN²=OP²-ON²

AN²=OA²-ON²

∴PN²-AN²=(OP²-ON²)-(OA²-ON²)

=>PN²-AN²=OP²-ON²-OA²+ON²

=>PN²-AN²=OP²-OA² (proved)

•

now...

PN²-AN²=OP²-OA²......(a)

and from the ∆POT(right angled triangle)

PT²=OP²-OA²..............(b)

from (a)and (b)...

∴PN²-AN²=PT². (proved)

$<marquee direction="Right">hope this help you$