NCERT Class 9 Solutions: Quadrilaterals (Chapter 8) Exercise 8.2 – Part 4

  • Corresponding Angles

    Hear m and n two parallel line ,∠1 = ∠5,∠2 = ∠6,∠4 =∠8,∠3 = ∠6

    Hear M and N Two Parallel Line

    Hear m and n two parallel line ,∠1 = ∠5,∠2 = ∠6,∠4 =∠8,∠3 = ∠6

    1=52=64=83=6

  • Corresponding Angle:-the angles which occupy the same relative position at each intersection where a straight line crosses two others. If the two lines are parallel, the corresponding angles are equal.

Q-7 ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that

  1. D is the mid-point of AC

  2. MDAC

  3. CM=MA=12AB

Solution:

Triangle poly1 Triangle poly1: Polygon B, C, A Triangle poly1 Triangle poly1: Polygon B, C, A Segment c Segment c: Segment [B, C] of Triangle poly1 Segment a Segment a: Segment [C, A] of Triangle poly1 Segment b Segment b: Segment [A, B] of Triangle poly1 Segment f Segment f: Segment [C, M] Segment g Segment g: Segment [M, D] Point B B = (-1.94, 4.32) Point B B = (-1.94, 4.32) Point B B = (-1.94, 4.32) Point C C = (-1.94, 0.22) Point C C = (-1.94, 0.22) Point C C = (-1.94, 0.22) Point A A = (2.62, 0.22) Point A A = (2.62, 0.22) Point A A = (2.62, 0.22) Point M Point M: Point on b Point M Point M: Point on b Point M Point M: Point on b Point D Point D: Point on a Point D Point D: Point on a Point D Point D: Point on a

ABC Is a Triangle

ABC is a triangle is the midpoint of AB and MM||BC and D is the midpoint of AC

  • Give triangle of ABC at C, a line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D.

  1. In ΔACB ,

    • M is the midpoint of AB and MD BC

    • Thus; D is the midpoint of AC (Converse of mid-point theorem)

  2. ACB=ADM (Corresponding angles)

    • Also, ACB=90°

    • Thus, ADM=90° and MDAC

  3. In ΔAMD and ΔCMD ,

  • AD=CD (D is the midpoint of side AC)

  • ADM=CDM (Each 90°)

  • DM=DM (Common)

  • Thus, ΔAMDΔCMD by Side-Angle-Side congruence condition.

  • AM=CM by Corresponding Part of Congruent Triangles

  • also, AM=12AB (M is mid-point of AB)

  • Hence, CM=MA=12AB

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