Question

Point E is in the exterior of a circle. A secant through E intersects the circle at points A and B and a tangent through E touches the circle at point T then prove that EA X EB = ET^2​

Answer 1

$\underline { \blue { Given: }}$

A secant EAB to a circle C(O,r) ,intersecting it in A and B and ET is a tangent segment.

$\underline { \blue { To \: Prove: }}$

$EA \times EB = ET^{2}$

$\underline { \blue { Construction: }}$

Draw OD Perpendicular to AB .

Join OE , OT and OA.

$\underline { \blue { Proof: }}$

Since the foot of the perpendicular from the centre to a chord bisects the chord .

OD perpendicular to AB

=> AD = DB ----(1)

Now,

$EA \times EB = ( ED - AD)(ED + DB)\\= (ED - AD)( ED + AD) \: [ From \: (1) ]$

$= ED^{2} - AD^{2}\\= ( OE^{2} - OD^{2}) - AD^{2}$

/* In right angle OPD , we have

OE² = OD² + ED² => ED² = OE² - OD² */

$OE^{2} - ( OD^{2} + AD^{2} )$

/* In right angle OAD , we have

OA² = OD² + AD² */

$= OE^{2} - OA^{2} \: [ Since, OA = OT = r ]$

$= OE^{2} - OT^{2}$

$( Since , \angle {OTE} = 90\degree , \\Therfore , OE^{2} = OT^{2} + ET^{2} \\ \implies OE^{2} - OT^{2} = PT^{2} )$

$= ET^{2}$

$Hence, EA \times EB = ET^{2}$

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