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- 1. By iTutor.com T- 1-855-694-8886 Email- info@iTutor.com
- 2. Angle Sum Property of a Quadrilateral The sum of the angles of a quadrilateral is 360º. Given: ABCD is a quadrilateral, To Prove: <A + <B + <C + <D = 360º Construction: Draw AC be a diagonal (see Fig.) Proof: In ∆ ADC we know that D A ∠ DAC + ∠ ACD + ∠ D = 180° ……………(i) C B
- 3. Angle Sum Property of a Quadrilateral Similarly, in ∆ ABC, ∠ CAB + ∠ ACB + ∠ B = 180° …………. (ii) Adding (i) and (ii), we get ∠ DAC + ∠ ACD + ∠ D + D ∠ CAB + ∠ ACB + ∠ B = 180° + 180° = 360° Also, ∠ DAC + ∠ CAB =∠ A and ∠ ACD + ∠ ACB = ∠ C So, i.e., C A B ∠ A + ∠ D + ∠ B + ∠ C = 360° the sum of the angles of a quadrilateral is 360°.
- 4. Properties of a Parallelogram A diagonal of a parallelogram divides it into two congruent triangles. Given: ABCD be a parallelogram and AC be a diagonal AC divides into two D C triangles, ∆ ABC & ∆ CDA. To prove: ∆ ABC ≅ ∆ CDA Proof: In ∆ ABC and ∆ CDA We know that BC || AD and AC is a transversal So, A ∠ BCA = ∠ DAC ….(Pair of alternate angles) B
- 5. Properties of a Parallelogram Also, So, AB || DC and AC is a transversal ∠ BAC = ∠ DCA ….........Pair of alternate angles And D C AC = CA …………… Common So, by the Rule of ASA ∆ ABC ≅ ∆ CDA A Diagonal AC divides parallelogram ABCD into two congruent triangles ABC and CDA B
- 6. Properties of a Parallelogram If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram. Given: Quadrilateral ABCD, sides AB and CD is equal and also AD = BC D C To Prove: ABCD is a Parallelogram Construction: Draw diagonal AC Proof: In ∆ ABC & ∆ CDA ∆ ABC ≅ ∆ CDA ……… S. S. S. So, ∠ BAC = ∠ DCA A B and ∠ BCA = ∠ DAC, So that the line AB||DC and AD||BC , So that ABCD is a parallelogram
- 7. Properties of a Parallelogram If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. Given: Quadrilateral ABCD, diagonal AC and BD Where OA = OC and OB = OD D C To Prove: ABCD is a parallelogram So, ∆ AOB ≅ ∆ COD ….….. SAS Therefore, O ∠ ABO = ∠ CDO From this, we get AB || CD Similarly, A BC || AD Therefore ABCD is a parallelogram. B
- 8. Properties of a Parallelogram A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel. Given: ABCD be a quadrilateral in which AB = CD & AB||CD. D C Construction: Draw diagonal AC Proof: In ∆ ABC & ∆ CDA AC = AC …………same AB = CD …………Given <BAC = < ACD …..…Given A B ∆ ABC ≅ ∆ CDA ………..by SAS congruence rule. So, BC || AD
- 9. Properties of a Parallelogram The opposite sides and opposite sides of a parallelogram are respectively equal in measure. b c 3 Given: Parallelogram abcd 4 1 a 2 To Prove: |ab| = |cd| and |ad| = |bc| and abc = adc d Construction: Draw the diagonal |ac|
- 10. Properties of a Parallelogram The opposite sides and opposite sides of a parallelogram are respectively equal in measure. b c 3 4 1 a Proof: In the triangle abc and the triangle adc 1 = 4 …….. Alternate angles 2 = 3 ……… Alternate angles 2 d |ac| = |ac| …… Common The triangle abc is congruent to the triangle adc ………ASA |ab| = |cd| and |ad| = |bc| and abc = adc = ASA.
- 11. Mid-point Theorem The line which joins the midpoints of two sides of a triangle is parallel to the third side and is equal to half of the length of the third side Given: In ∆ ABC where E and F are mid-points of AB and AC respectively To Prove: (i) EF || BC A E (ii) EF = 1 BC 2 B F C
- 12. Mid-point Theorem Construction: Draw DC || AB to meet EF produced at D. A Proof: In ∆ AEF & ∆ CDF EFA = CFD .................V. A. A. ………………..Given AF = CF AEF = CDF E ∆ AEF ≅ ∆ CDF And D …............A. I. A. B So, F ………..…..A. S. A. Rule EF = DF and BE = AE = DC DC || BE , the quadrilateral BCDE C
- 13. Mid-point Theorem Therefore, BCDE is a parallelogram from the Properties of parallelogram A gives EF || BC. ---------------(i) Proved E F D BC = DE = EF + FD We Know that EF = DE So, BC = EF + EF Or, B BC = 2 EF Or , EF = 1 2 BC -------------(ii) Proved C
- 14. Mid-point Theorem The line through the midpoint of one side of a triangle when drawn parallel to a second side bisects the third side. Given: In ∆ ABC where E is the mid-point of AB, line l is passing through E and is parallel to BC . A To Prove: AF = FC E B F l C
- 15. Mid-point Theorem Construction: Draw a line CM || AB to meet EF produced at D. A M Proof: CM || AB ………..(Const.) E F D EF || BC …………(Given) B So, Quadrilateral BCDE is a parallelogram, then BE = CD Now In ∆ AEF and ∆ CDF. CD = BE = AE , C
- 16. Mid-point Theorem CFD = EFA …………..(Vertically apposite) DCF = EAF A …..(Alternate) So, ∆ AEF ≅ ∆ CDF …(ASA Rule) AF = CF Proved . B E M F D C
- 17. Areas Of Parallelograms Parallelograms on the same base and between the same parallels are equal in area. Given: Two parallelograms ABCD and EFCD, A E On the same base DC and between the same parallels AF and DC To Prove: D ar (ABCD) = ar (EFCD) B C F
- 18. Areas Of Parallelograms Proof: In ∆ ADE and ∆ BCF, AD || BC ∠ DAE = ∠ CBF ……….. (Corresponding angles) A ED || FC ∠ AED = ∠ BFC E B (Corresponding angles) Therefore, ∠ ADE = ∠ BCF D C (Angle sum property of a triangle) Also, So, AD = BC (Opposite sides of the parallelogram ABCD) ∆ ADE ≅ ∆ BCF ………(By ASA rule) F
- 19. Areas Of Parallelograms Therefore, ar (ADE) = ar (BCF) -------(Congruent figures have equal areas) ar (ABCD) = ar (ADE) + ar (EDCB) A E B F = ar (BCF) + ar (EDCB) = ar (EFCD) So, D C Parallelograms ABCD and EFCD are equal in area.
- 20. Areas Of Triangles Two triangles on the same base and between the same parallels are equal in area. A P Given: ∆ ABC and ∆ PBC on the same base BC and between the same parallels BC and AP B To Prove: ar (ABC) = ar (PBC) C
- 21. Areas Of Triangles Construction: Draw CD || BA and CR || BP such that D and R lie on line AP A P D R Proof: From this, B C We have two parallelograms PBCR and ABCD on the same base BC and between the same parallels BC and AR. So, ar (ABCD) = ar (PBCR) ∆ ABC ≅ ∆ CDA and ∆ PBC ≅ ∆ CRP (A diagonal of a parallelogram divides it into two congruent triangles)
- 22. Areas Of Triangles So, 1 ar (ABC) = 2 ar (ABCD) A D P (AC is a diagonal of ABCD) And B 1 ar (PBC) = 2 ar (PBCR) (PC is the diagonal of PBCR) Therefore, ar (ABC) = ar (PBC) Proved C R
- 23. Areas Of Triangles Area of a triangle is half the product of its base (or any side) and the corresponding altitude (or height) Proof : Now, suppose ABCD is a parallelogram whose one of the diagonals is AC, Draw AN ⊥ DC. D So, B A N C ∆ ADC ≅ ∆ CBA …………(ABCS is a parallelogram) ar (ADC) = ar (CBA). …(ABCS is a parallelogram)
- 24. Areas Of Triangles Therefore, ar (ADC) = 1 ar (ABCD) 2 1 = 2 B A (DC× AN) D N C area of ∆ ADC = 1 × base DC × corresponding altitude AN 2 In other words, Area of a triangle is half the product of its base and the corresponding altitude.
- 25. Call us for more Information: 1-855-694-8886 Visit www.iTutor.com The End

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