Question

9. In Fig. 10.13, XY and X'Y' are two parallel tangents to a circle with centre 0 and
another tangent AB with point of contact Cintersecting XY at A and X'Y' at B. Prove
that 2 AOB = 90.


Figure is attached.. Plz help... No spam​

Answer 1

${}{ \huge{ \underline{ \red{answer - }}}}$

⇒Given that :

XY is tangent to the circle at point P.

X’Y’ is tangent to the circle at point Q.

AB is tangent to the circle at point C.

and XY ∥ X’Y’

⇒To prove :

∠AOB = 90°

⇒Proof :

First join OC such that OC is perpendicular to AB.

In ΔOPA and ΔOCA,

OP = OC (Radius of the same circle)

AP = AC (Tangents from point A)

AO = AO (Common side)

⇒ΔOPA ≅ ΔOCA (SSS congruence criterion)

∠POA = ∠COA … (i)

Similarly, ΔOQB ≅ ΔOCB

∠QOB = ∠COB … (ii)

Since POQ is a diameter of the circle, it is a straight line.

Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180°

From equations (i) and (ii), it can be observed that

⇒ 2∠COA + 2 ∠COB = 180°

⇒∠ COA + ∠ COB = 180° / 2

⇒∠COA + ∠COB = 90°

⇒∠AOB = 90°

Hence Proved.

=============================!!

Answer 2

$\large{\underline{\underline{\mathfrak{\sf{\red{Correct\:Question-}}}}}}$

XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that ∠AOB = 90.

$\large{\underline{\underline{\mathfrak{\sf{\red{Explanation-}}}}}}$

Given :

• XY and X'Y' are two parallel tangents to a circle with centre 0 and another tangent AB with point of contact Cintersecting XY at A and X'Y' at B.

To prove :

• ∠AOB = 90

Construction :

• Join OC

Proof :

In ∆AOP and ∆AOC,

• OA = OA ( Common side )
• AP = AP ( Tangents from an external point are equal in length )
• OP = OC ( Radii of same circle are equal )

$\therefore$ ∆AOP ≅ ∆AOC ( By SSS rule )

$\implies$ ∠AOP = ∠AOC ( Corresponding parts of congruent ∆s are equal )⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀–eq (1)

$\rule{200}2$

Similarly, in ∆BOC and ∆BOQ,

• OB = OB ( Common side )
• BC = BQ ( Tangents from an external point are equal in length )
• OC = OQ ( Radii of same circle are equal )

$\therefore$ ∆BOC ≅ ∆BOQ ( By SSS rule )

$\implies$ ∠BOC = ∠BOQ ( Corresponding parts of congruent ∆s are equal )⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀–eq (2)

$\rule{200}2$

Now, ∠AOP + ∠AOC + ∠BOC + ∠BOQ = 180° ( Sum of angles on a straight line is of 180° )

★ Using (1) and (2),

$\implies$ ∠AOC + ∠AOC + ∠BOC + ∠BOC = 180°

$\implies$ 2∠AOC + 2∠BOC = 180°

$\implies$ 2(∠AOC + ∠BOC) = 180°

$\implies$ ∠AOB = $\sf{\cancel{\dfrac{180}{2}}}$

$\implies$ $\large{\underline{\boxed{\mathfrak{\sf{\green{∠AOB=90°}}}}}}$

Hence proved!

$\rule{200}2$

#answerwithquality

#BAL