The line drawn through the center of a circle to bisect a chord is perpendicular to the chord. Given: A circle with center O, AB is chord of a circle and OC bisect chord at C. i,e. AC = CB. To prove: OC ⊥ AB Construction: join OA and OB. Proof: In △ OCA and △ OCB OA = OB (Radius […]
Read More →The perpendicular from the center of a circle to a chord bisects the chord. Given: A circle with center O, AB is chord of a circle and OC perpendicular from the center O to the chord AB. i.e. OC ⊥ AB therefore ∠OCA and ∠OCB, Both angles are 900. To prove: AC = CB Construction: join OA and OB. Proof: In △OCA and △ OCB ∠OCA = ∠OCB […]
Read More →If the angles subtended by chords of a circle at the center are equal, then the chords are equal. Given: A circle with center O, AB and CD are chords of circle that subtend equal angles at center O. i,e. ∠AOB = ∠COD. To prove: chord AB = chord CD Proof: In △AOB and △ CODOA = OC (Radius of circle)∠AOB […]
Read More →Equal chords of a circle subtend equal angles at the center. Given: Two equal chords AB and CD of a circle with center O. i,e. AB = CD. To prove: ∠AOB = ∠COD Proof: In △AOB and △ COD OA = OC (Radius of circle)OB = OD (Radius of circle)AB = CD (Given) Hence, △AOB ≅ △ COD (SSS Congruence rule) ∴ ∠AOB […]
Read More →A diagonal of a parallelogram, divides it into two congruent triangles. Given: A parallelogram ABCD and AC is a diagonal, the diagonal AC divides parallelogram ABCD into two triangles △ ABC and △ CDA. To prove: These triangles △ ABD and △CDA are congruent, △ ABC ≅ △ CDA Proof: In △ ABC and △ CDA BC ∥ AD and AC […]
Read More →In a parallelogram opposite sides are equal. Given: A parallelogram ABCD, each pair of Opposite sides of parallelogram are side AB and side DC and side AD and side BC. To prove: Opposite sides of parallelogram are equal that is AB = DC and AD = BC Construction: Join A to C that is AC, is a diagonal, the diagonal AC […]
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