Question
Sat February 14, 2009 By:

# prove

Sat February 14, 2009

Consider two parallel tangents AB and CD  where A and C are the points of contact of the tangents with the circle.

Let BD be the portion which is the intercept of another tangent between the two parallel tangents AB and CDand E be the point of contact of that tangent portion with the circle.

Let O be the centre of the circle.

Consider triangle AOB and EOB.

AB=BE (tangent segments drawn from an external point are equal in length.)

angle OAB=angle OEB= 90 degrees each..9 radius thru' the point of contact is perpendicular to the tangent)

So,

triangle OAB is congruent to triangle OEB  (RHS rule)

So,

angle AOB= angle EOB (C.P.C.T.)=x ( say)

Similarly,

angle COD = angle EOD = y( say)

but angle AOC is =180 degrees ( straight angle)

So,

angle AOB+ angle EOB+ angle EOD+ angle COD= 180

So,

2x+2y=180

So,

x+y=90

So,

angle BOE+ angle EOD= 90

So,

angle BOD =90

Hence proved.

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