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Mathematics in Daily Life 9th Grade Theorems on Parallelograms Objective After learning this chapter, you should be able to • • • • Prove the properties of parallelograms logically. Explain the meaning of corollary. State the corollaries of the theorems. Solve problems and riders based on the theorem. 2 Flowchart on Procedure to Prove a Theorem Let us recall the procedure of proving a theorem logically. Observe the following flow chart. Consider/take a statement or the Enunciation of the theorem For example, in any triangle the sum of three angles is 180˚ A Draw the appropriate figure and name it. B C Write the data using symbols. ABC is a triangle 1 2 3 Flowchart on Procedure to Prove a Theorem 1 2 Write what is to be proved using symbols BAC ABC BCA 180 Analyze the statement of the theorem and write the hypothetical construction if needed and write it symbolically Through the Vertex A draw EF || BC E A F Write the reason for construction Draw the appropriate figure and name it. Use postulates, definitions and previously proved theorems along with what is given and construction 4 Theorems on Parallelograms Theorem 1: The diagonals of a parallelogram bisect each other. Theorem 2: Each diagonal divides the parallelogram into two congruent triangles. 5 Theorem 1 Proof Theorem: The diagonals of a parallelogram bisect each other. D C O A B Given: ABCD is a parallelogram. AC and BD are the diagonals intersecting at O. To Prove: AO = OC BO = OD 6 Theorem 1 Proof Contd.. Proof: Statement Reasons Process of Analysis 1) In ∆AOB and ∆COD, AB = DC 2) AOB COD Opposite sides of the parallelogram Vertically opposite angles Recognise the ∆s which contain the sides AO, BO, CO, DO. 3) OAB OCD ABO ODC Alternate angles AB || DC and BD is a transversal. ASA Postulate Use the data to prove the congruency of these two ∆s ΔAOB ΔCOD AO OC and BO OD Corresponding sides of congruent ∆s i.e., The diagonals of parallelogram bisect each other. 7 Theorem 2 Proof Theorem: Each diagonal divides the parallelogram into two congruent triangles. D C O A B Given: ABCD is a parallelogram in which AC is a diagonal. AC = DC, AD = BC To Prove: ΔABC ΔADC 8 Theorem 2 Proof Contd.. Proof: Statement 1) AB = DC Reasons Opposite sides of the parallelogram 2) BC = AD Opposite sides of the parallelogram 3) AC is common ΔABC ΔADC S.S.S. postulate Diagonal AC divides the parallelogram ABCD into two congruent triangles. Similarly, we can prove that ΔABD ΔBDC. Each diagonal divides the parallelogram into two congruent triangles. 9 Corollary A corollary is a proposition that follows directly from a theorem or from accepted statements such as definitions. Corollaries of the Theorems There are four corollaries for the theorems explained in the previous slides. They are, Corollary-1: In a parallelogram, if one angle is a right angle, then it is a rectangle. 10 Corollaries of the Theorems Contd… Corollary-2: In a parallelogram, if all the sides are equal and all the angles are equal, then it is a square. Corollary-3: The diagonals of a square are equal and bisect each other perpendicularly. Corollary-4: The straight line segments joining the extremities of two equal and parallel line segments on the same side are equal and parallel. 11 Corollary 1 Proof Corollary: In a parallelogram, if one angle is a right angle, then it is a rectangle. R S 90˚ P Q o Q 90 . Given: PQRS is a parallelogram. Let To Prove: PQRS is a rectangle. 12 Corollary 1 Proof Contd.. Proof: Statement Q 90o Given ---- (1) Opposite angles of parallelogram PQRS Q S 90o Q R 180 o 90o R 180o R 180o 900 R 90o R P 90o Reasons Sum of the consecutive angles of a parallelogram PQRS is equal to 180˚ By substitution By transposing ----(2) ----(3) Opposite angles of parallelogram PQRS P Q R S 90o From the equations (1), (2) and (3) Hence, PQRS is a rectangle. 13 Corollary 2 Proof Corollary: In a parallelogram, if all the sides are equal and all the angles are equal, then it is a square. D A C 90˚ 90˚ 90˚ 90˚ B Activity!!! Prove this corollary logically 14 Corollary 3 Proof Corollary: The diagonals of a square are equal and bisect each other perpendicularly. D C 90˚ 90˚ 90˚ 90˚ A B Given: ABCD is a square. AB = BC = CD = DA. A B C D 90o To Prove: 1) AC = BD 2) AO = CO, BO = DO. 3) AOB DOC 90o 15 Proof: Corollary 3 Proof Contd.. Statement Reasons 1) Consider the ∆ABD and ∆ABC AB = BC Sides of the square are equal Angles of the square are equal BAD ABC 90o AB is common. S.A.S postulate ΔABD ΔABC Congruent parts of congruent ∆s AC BD 2) Consider the ∆AOB and ∆DOC AB = DC Opposite sides of the square OAB OCD Alternate angles AB || DC ABO CDO Alternate angles AB || DC A.S.A postulate ΔAOB ΔDOC Congruent parts of congruent ∆s AO CO, BO DO Hence, diagonals of a square bisect each other 16 Corollary 3 Proof Contd.. Statement 3) Consider the ∆AOD and ∆COD AD = CD AO = CO DO is common ΔAOD ΔCOD AOD COD AOD COD 180o Reasons Sides of the square are equal The diagonals bisect each other S.S.S postulate Congruent parts of congruent ∆s Linear pair AOD COD 90o Hence, the digonals bisect each other at right angles. 17 Corollary 4 Proof Corollary: The straight line segments joining the extremities of two equal and parallel line segments on the same side are equal and parallel. Activity!!! Prove this corollary logically Hint :- S.A.S. Postulate of congruency triangles 18 Examples Example-1: In the given figure, ABCD is a parallelogram in which DAB 70o and DBC 80o . Calculate the angles CDB and ADB. D Given: ABCD is a parallelogram AB = DC, AD = BC AB || DC, AD || BC DAB 70 o C 80˚ and DBC 80 . A o 70˚ B To Find: CDB and ADB 19 Examples Contd... Solution: Statement DAB 70 o BCD Reasons Opposite angles of the parallelogram ABCD. In ∆BDC, DBC BCD CDB 180o 80o 70o CDB 180o CDB 180o 150o CDB 30 o DBC ADB Sum of three angles of a triangle By substitution By transposing Alternate angles, AD || BC. ADB 800 20 Examples Contd… Example-2: In the figure, ABCD is a parallelogram. P is the mid point of BC. Prove that AB = BQ. D Given: ABCD is a parallelogram ‘P’ is the mid point of BC. BP = PC A C P Q B To Prove: AB = BQ 21 Examples Contd... Solution: Statement Consider the ∆BPQ and ∆CPD, BP = PC BPQ CPD PBQ PCD ΔBPQ ΔCPD BQ DC ------(1) But DC = AB BQ AB ------(2) Reasons Given Vertically opposite angles Alternate angles, AB || DC A.S.A. postulate C.P.C.T Opposite sides of the parallelogram From (1) and (2) 22 Exercises 1. In a parallelogram ABCD,A =60⁰. If the bisectors of A and B meet at P on DC. Prove that APB 90o. 2. In a parallelogram ABCD, X is the mid-point of AB and Y is the mid-point of DC. Prove that BYDX is a parallelogram. 3. If the diagonals PR and QS of a parallelogram PQRS are equal, prove that PQRS is a rectangle. 4. PQRS is a parallelogram. PS is produced to M so that SM = SR and MR is produced to meet PQ produced at N. prove that QN = QR. 5. ABCD is a parallelogram. If AB = 2 x AD and P is the mid-point of AB, prove that CPD 90 o. 23