Question

Prove that the length of tangents drawn from an external point to a circle are equal

Answer 1

Let two tangent PT and QT are drawn to circle of centre O as shown in figure.
Both the given tangents PT and QT touch to the circle at P and Q respectively.

We have to proof : length of PT = length of QT
Construction :- draw a line segment ,from centre O to external point T { touching point of two tangents } .

Now ∆POT and ∆QOT
We know, tangent makes right angle with radius of circle.
Here, PO and QO are radii . So, ∠OPT = ∠OQT = 90°
Now, it is clear that both the triangles ∆POT and QOT are right angled triangle.
nd a common hypotenuse OT of these [ as shown in figure ]

Now, come to the concept ,
∆POT and ∆QOT
∠OPT = OQT = 90°
Common hypotenuse OT
And OP = OQ [ OP and OQ are radii]
So, R - H - S rule of similarity
∆POT ~ ∆QOT
Hence, OP/OQ = PT/QT = OT/OT
PT/QT = 1
PT = QT [ hence proved]

Answer 2

Given:-

Tangent AB and AC are on points B and C.

$\bf{To \: prove}$

AB=AC

$\bf{ \blue{construction}}$

join O to A

O to C

and

O to B

$\bf{ \purple{proof}}$

$\sf{in \triangle \: aoc \: and \: \triangle \: aob}$

$\sf{ \angle \: oca = \angle \: oba( \: both \: 90 \degree)}$

$\sf{ob = oc \: \: (radius \: of \: circle)}$

$\sf{oa = oa \: (common)}$

$\sf{ hence\: \triangle \: aoc \: \cong\: \triangle \: aob(by \: rhs)}$

Hence

$\bf{ab = ac \: \: (cpct)}$

$\huge{LHS\:=\:RHS}$

$\bf{\underline{proved}}$