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NCERT Class 9 Solutions: Triangles (Chapter 7) Exercise 7.1 – Part 4

Illustration: NCERT Class 9 Solutions: Triangles (Chapter 7) Exercise 7.1 – Part 4

When two lines are cut by a third line (transversal) co-interior angles are between the pair of lines on the same side of the transversal. If the lines are parallel the co-interior angles are supplementary (add up to 180°) .

Q-7 AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that and (see Fig) . Show that

Ab is a Line Segment and P is Its Mid-Point

Solution:

Given, P is mid-point of AB.

  1. (Adding both sides)

    In ,

    (P is mid-point of AB)

    (Given)

    Therefore, By Angle-Side-Angle congruence condition.

  2. (By corresponding Part of Congruent Triangles) .

Q-8 In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see Fig) . Show that:

  1. ΔAMC ≅ ΔBMD
  2. ∠ DBC is a right angle.
  3. ΔDBC ≅ ΔACB
  4. CM = 1⟋2 AB
Right Triangle ABC

Solution:

Given, right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to point D

M is the mid-point of AB and

  1. In and

    (M is the mid-point)

    (Vertically opposite angles)

    (Given)

    Therefore, (By Side-Angle-Side congruence condition) .

  2. (By Corresponding Parts of Congruent Triangles) Therefore,

    as alternate interior angles are equal.

    Now,

    (co-interiors angles)

    In ΔDBC and ΔACB,

    (Common)

    (Right angles)

    (By Corresponding Parts of Congruent Triangles, already proved)

    Therefore (By Side-Angle-Side congruence condition)

  3. (M is mid-point)