Question

class 9 theoram 10.4

Answer 1

$\mathbb{ELLO \ THERE}$

Theorem 10.4

A line drawn through the center of a circle to bisect a chord is perpendicular to the chord.

$\underline{\mathsf{ANSWER}}$

$\textsf{Given}$

AB is a chord in the circle, with centre O.

AC = BC

$\textsf{To \ Prove}$

OA ⊥ BA

$\textsf{Construction}$

Join OA and BO

$\textsf{Proof}$

In ΔAOC and ΔBOC

AC = BC (given)

AO = BO (radii of the same circle)

OC = OC (common)

∴ ΔAOC ≅ ΔBOC by SSS congruency

∠OCA = ∠OCB (CPCT)

Now,

∠OCA + ∠OCB = 180°  (Linear Pair)

2∠OCB = 180°

 ∠OCB = $\frac{180}{2}$

 ∠OCB = 90°

Since

∠OCA = ∠OCB

∠OCA = ∠OCB = 90°

⇒ OA ⊥ BA

(Angles on either side are 90)

Hence Proved!

$\large\boxed{\large\boxed{\mathbb{THANK \ YOU}}}$

Answer 2

Step-by-step explanation:

The line drawn through the centre of the circle to bisect a chord is perpendicular to the chord.

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