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

m and n two parallel lines cut by a transversal: ∠1 = ∠5, ∠2 = ∠6, ∠4 =∠8, ∠3 = ∠6

M and N Are Two Parallel Line Cut by a Transversal

m and n two parallel lines cut by a transversal: ∠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 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 Right Triangle

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

Given

  • Triangle of ABC with right angle at C.

  • 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 MDBC

    • 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.

Thus, AM=CM (Corresponding Parts of Congruent Triangles)

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

Therefore, CM=MA=12AB

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