Question

explain and proof alternate segment theorem pls answer karo ye 3 rd time post kr rha hu question​

Answer 1

Alternate Segment Theorem :- The Angle b/w a chord and a tangent through one of the end points of the chord is equal to the angle in the alternate segment.

Proof :-

From image we have :-

→ AB = A Chord .

→ AF = Tangent to the circle.

→ O = centre .

→ OB = OA = radius of circle.

→ ∠BAF = Let x° .

To Prove :-

→ ∠ACB = x° ..

Solution :-

in ∆AOB,

→ OB = OA = radius .

So,

→ ∠OAB = ∠OBA . (Angles opposite to equal sides of a triangle are equal.)

Than,

→ ∠OAB + ∠OBA + ∠AOB = 180° (Angle sum Property.)

→ ∠OAB + ∠OAB + ∠AOB = 180°

→ 2∠OAB + ∠AOB = 180°

→ ∠AOB = (180° - 2∠OAB) ----------- Eqn.(1)

______________

Now,

→ AF = Tangent to the circle.

So,

→ ∠OAF = 90° (A tangent to a circle forms a right angle with the circle's radius).

So,

→ x° = ∠OAF - ∠OAB

→ x° = (90° - ∠OAB)

→ ∠OAB = (90° - x) -------------- Eqn.(2) .

______________

Putting value of Eqn.(2) in Eqn.(1) Now,

→ ∠AOB = 180° - 2(90° - x)

→ ∠AOB = 180° - 180° + 2x

→ ∠AOB = 2x .

Now, we know That, The angle subtended by an arc at the center of a circle is double that of the angle that the arc subtends at any other given point on the circle.

Therefore, if ,

→ ∠AOB = 2x (At centre.)

Than,

→ ∠ACB = (2x°/2) = x° (At circumference.)

Hence,

→ ∠ACB = x° . (Proved).

___________________________

Extra :-

→ ∠CAE = ∠CBA . (By Alternate Segment Theorem.)

___________________________

Answer 2

${ \huge{ \underline{ \underline{ \mathfrak{ \red{question:-}}}}}}$

▪ Explain and Prove the alternate segment theorem.

${ \huge{ \underline{ \underline{ \mathfrak{ \red{solution:-}}}}}}$

${ \pink{ \bold{ \underline{alternate \: segment \: theorem}}}}$

For any circle, the angle between a tangent and a chord through the point of contact of the tangent is equal to the alternate segment.

${ \boxed{ \underline{ \bold{ \purple{proof-}}}}}$

▪ In the attached figure ,

${ \bold{ \underline{ \purple{given-}}}}$

¤ O is the centre of the circle

¤ AB is a chord of the circle

¤ OA and OB are the radii of the circle

¤ XY is a tangent passing through point A on the circle.

▪ Let ∠ BAY be x°.....

${ \bold{ \underline{ \purple{to \: proof-}}}}$

☆ ∠BCA = ∠BAY = x°

${ \bold{ \pink{ \: in \: triangle \: AOB }}}$

▪ OA = OB ( radius of circle )

then,

▪ ∠OAB = ∠OBA

( angles opposite to equal sides are also equal)

using,

ANGLE SUM PROPERTY of triangle....

${ \boxed{ \bold{ \red{sum \: of \: all \: 3 \: angles = 180}}}}$

∠AOB + ∠OAB + ∠OBA = 180°

=》 ∠AOB + ∠OAB + ∠OAB = 180°

=》 ∠AOB + 2∠OAB = 180°

=》 ∠AOB = (180° - 2∠OAB )------> eqn (1)

___________________________________

∠OAY = 90°

( since , XY is a tangent to the circle, it forms right angle with the radius of the circle)

∠OAY = ∠OAB + ∠BAY = 90°

=》 ∠OAB + x° = 90°

=》 ∠OAB = ( 90° - x° )-------> eqn (2)

substituting the value of ∠OAB from equation (2) in equation (1)......

∠AOB = ( 180° - 2∠OAB)

=》 ∠AOB = ( 180° - 2 ( 90° - x° ))

=》 ∠AOB = 180° - 180° + 2x°

=》 ∠AOB = 2x°

since,

The angle subtended by an arc at the centre of a circle is double of the angle that the arc subtends at any given point in the circle.....

thus,

∠AOB = 2 ∠BCA

=》 ∠BCA = ∠AOB/ 2

=》 ∠BCA = 2x°/2 = x°

therefore,

∠BAY = ∠BCA = x°

.......Hence, PROVED