Question

Prove that the intercept of a tangent between two parallel tangents to circke subtends a right angle at the centre.

Answer 1

prove that the intercept of a tangent between two parallel tangents to acircle subtends a right angle triangle at the centre. Given: XY and X'Y' at are two parallel tangents to thecircle wth centre O and AB is thetangent at the point C,which intersects XY at A and X'Y' at B.

Answer 2

$\huge\sf\mathbb\color{white} \underline{\colorbox{black}{⚔☯☃️SoLuTiOn☯⚔☃️}}$

Step-by-step explanation:

Let XY and X'Y' are two parallel tangents and AB is intercept of another tangent. O is centre of circle.

Since, the length of the tangent drawn from external point are equal, hence, as the figure,

$\: \: \: \: \: \: \: \: \: \: \: \: AP=AC$

$∴ \: \: In \: ΔAPO \: and \: ΔACO,</p><p> \\ AP=AC$

$\: \: \: \: \: \: \: \: \: \: \: AO=AO, \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( common)$

$\: \: \: \: \: \: \: \: \: \: \: OP=OC, \: \: \: \: \: \: \: (radii \: of \: the \: circle)$

$∴ \: \: \: ΔAPO≅ΔACO, \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( SSS congruency)$

$⟹ \: \: \: ∠PAO=∠OAC$

$⟹ \: \: \: \: ∠PAO=2∠OAC \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2∠OAC$

Similarly we can prove that,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ∠CBO=∠OBQ$

$⟹ \: \: \: \: \: \: \: \: ∠CBQ=2∠CBO$

Since, XY || X'Y'

$∴ \: \: \: \: ∠PAC +∠QBC=180ﾟ, \\ (interior \: angles \: of \: the \: same \: side)$

$⟹2∠CAO+2∠CBO=180ﾟ \\ </p><p>⟹∠CAO+∠CBO=90ﾟ \: \: \: \: \: ...(1)$

$In \: ΔAOB,$

$∠CAO+∠CBO+∠AOB=180ﾟ$

$⟹ \: \: ∠CAO+∠CBO=180ﾟ-∠AOB$

$⟹ \: \: \: \: \: 90ﾟ=180ﾟ-∠AOB \\ [from \: eqn.(1)]$

$⟹180ﾟ - 90ﾟ=∠AOB$

$∴ \: \: \: \: \: \: \: \: \: \: \: \: ∠AOB=180ﾟ-90ﾟ$

Thus, intercept of another tangent subtends right angle at center.