Question

the lengths of tangent drawn from an external point to a circle are equal. prove it​

Answer 1

Answer:

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Answer 2

$\huge\mathfrak\pink{Answer}$

★${\sf{\underline{\underline{\pink{Given:-}}}}}$

Let Circle be with center O & P be a point outside circle PQ & PR are two tangents to circle intersecting at point Q & R respectively.

★${\sf{\underline{\underline{\pink{To \ prove:-}}}}}$

Lengths of tangents are equal

i.e. PQ & PR.

★${\sf{\underline{\underline{\pink{Construction:-}}}}}$

join OQ, OR & OP

★${\sf{\underline{\underline{\pink{Proof:-}}}}}$

As PQ is a tangent

OQ ⊥ PQ $\large\displaystyle{\boxed{\sf{\displaystyle{(Tangent \ of \ any \ point \ of circle \ is perpendicular \ to \ the \ ratio \ through \ points \ of \ contact)}}}}$

So, ∠OQP = 90°

Hence ▲OQP is right triangle

$\sf\red{Similarly}$

PR is a tangent

& OR ⊥ PR $\large\displaystyle{\boxed{\sf{\displaystyle{(Tangent \ of \ any \ point \ of circle \ is perpendicular \ to \ the \ ratio \ through \ points \ of \ contact)}}}}$

So, ∠ORP = 90°

Hence ▲ ORP is a triangle

${\sf{\underline{\underline{\pink{Using \ Pythagoras \ Theorm:-}}}}}$

$(Hypotenuse)^{2} \: = (Height)^{2} +( Base)^{2}$

In right angled triangle ▲OQP

$OP² = PQ² + OQ²$

$OP² - OQ² = PQ²$

$PQ² = QP² - OQ² --------(1)$

In right angled triangle ▲ORP

$OP² = PR² + OR²$

$OP² = PR² + OQ² (As \: OQ \: = OR \: both \: radius)$

$OP² - OQ² = PR²$

$PR² = OP² - OQ² ---------(2)$

★ from (1) & (2)

$PQ² = PR²$

$PQ = PR$

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