Question

prove that the length of two tangents drawn from an external point to circle are equal​

Answer 1

Answer:

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Given : AB and BC are two tangents mating to external point

To Prove : AB = BC

To Proof :

Method 1

In $\triangle{AOB} \: & \: \triangle{BOC}$

AO = OC...... (radius of circle)

OB = OB...... (same side)

$\angle{OAB} \: =\: \angle{BOC}$

Tangents meets the radius at 90°

$\triangle{AOB} \: \cong \: \triangle{BOC}$

By C.P.C.T ( Corresponding Parts Of Congruent Triangles)

$\overline{AB} \: =\: \overline{BC}$

Method 2

In $\triangle{AOB} \: & \: \triangle{BOC}$

$AB^{2}\: +\: AO^{2}\: =\: OB^{2}$

Eq. 1

$BC^{2}\: +\: CO^{2}\: =\: OB^{2}$

Eq. 2

Equation 1 & 2

$AB^{2}\: +\: AO^{2}\: =\: BC^{2}\: +\: CO^{2}$

AO = CO.. (radius of circle)

$AB^{2}\: +\: \cancel{CO^{2}} \: =\: BC^{2}\: +\: \cancel{CO^{2}}$

$AB^{2}\: =\: BC^{2}$

$\overline{AB} \: =\: \overline{BC}$

Hence Proved