Given: Two triangles △ ABC and △ ABD are on same base (or equal bases) AB, and area of △ ABC and △ ABD are equal. Proof: CD ∥ AB Construction: Draw CE and DF perpendicular to AB.So DF is the height of △ ADB, and CE is the height of △ABC. CE perpendicular to AB and DF perpendicular to AB.Since, lines perpendicular to same line […]
Read More →Two triangles on the same base (or equal bases) and between the same parallels are equal in area. Given: △ ABC and △ PBC are two triangles on same base (or equal bases) BC and between the same parallels, BC and AP. To prove: ar △ ABC = ar △ PBC Construction: Through B, draw BD ∥ CA intersecting PA produced at D, and […]
Read More →The sum of the angles of a triangle is 180º Given: A triangle △ PQR and ∠1, ∠2, and ∠3 are angles of triangle △ PQR. To prove: ∠1 + ∠2 + ∠3 = 1800 Construction: Draw a line XPY parallel to QR passing through P. Proof: XPY ∥ QR and PQ is transversal ∴ ∠2 = ∠4 (Alternate angles) ….(1) XPY ∥ QR and PR […]
Read More →Theorem – The line segment joining the mid-points of two sides of a triangle is parallel to the third side. Given: A triangle △ABC, in which D and E are mid points of sides AB and AC respectively. DE is joined. To prove: Line joining the mid points D and E (DE) is parallel to the third line […]
Read More →Theorem – Alternate Interior Angles are Equal If a transversal intersects two parallel lines, then each pair of alternate interior angles are equal. Given: Two parallel Lines AB and CD, and PS be transversal intersecting AB at Q and CD at R. To Prove: Each pair of alternate interior angles are equal. i.e. ∠BQR = ∠CRQ and ∠AQR = ∠QRD […]
Read More →Theorem 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 to the chord AB. i.e. OC ⊥ AB therefore ∠OCA and ∠OCB Both angles are 900. To prove: AC = CB (C is the mid point of chord AB) […]
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 a transversal intersect two parallel lines, then each pair of interior angles on the Same side of the transversal are Supplementary. Given: Two parallel Lines AB and CD, and PS be transversal intersecting AB at Q and CD at R. To Prove: Sum of interior angles on the Same side of the transversal is Supplementary. […]
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 →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 […]
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