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

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Co-interior angles

Give co-interior angles at two angle a and b

Give Co-Interior Angles

Give co-interior angles at two angle a and b

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 degrees).

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.

Ab is a line segment and P is its mid-point.D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB

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

right triangle ABC,right angled at C,M is the mid-point of hypotenuse AB.C is joind to M and produced to a point D such that DM = CM. Point D is joined to point B.

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)

  3. In ΔDBC and ΔACB, (Common) (Right angles) (By Corresponding Parts of Congruent Triangles, already proved)

    Therefore (By Side-Angle-Side congruence condition)

  4. (M is mid-point)