Class 9 theoram 10.4

1.	Class 9 theoram 10.4
Answer» $\mathbb{ELLO \ THERE}$ 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»

$\mathbb{ELLO \ THERE}$