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

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 Equation and Equation (see Fig). Show that

  1. Equation

  2. Equation

Angle α Angle α: Angle between D, A, F Angle β Angle β: Angle between E, B, F Angle γ Angle γ: Angle between A, F, E Angle γ Angle γ: Angle between A, F, E Angle δ Angle δ: Angle between D, F, B Angle δ Angle δ: Angle between D, F, B Segment f Segment f: Segment [A, B] Segment g Segment g: Segment [A, D] Segment h Segment h: Segment [B, E] Segment i Segment i: Segment [F, E] Segment k Segment k: Segment [D, G] Segment j Segment j: Segment [F, D] A text1 = "A" B text2 = "B" P text3 = "P" D text4 = "D" E text5 = "E"

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. Equation

  1. Equation (Adding Equation both sides) Equation Equation

    In Equation , Equation Equation (P is mid-point of AB) Equation (Given)Therefore, Equation (By Angle-Side-Angle congruence condition.

  2. Equation (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

Segment f Segment f: Segment [A, B] Segment g Segment g: Segment [B, C] Segment h Segment h: Segment [C, D] Segment i Segment i: Segment [D, B] Segment j Segment j: Segment [A, C] A text1 = "A" B text2 = "B" C text3 = "C" D text4 = "D" M text5 = "M"

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

Equation M is the mid-point of AB and Equation

  1. In Equation and Equation Equation (M is the mid-point) Equation (Vertically opposite angles) Equation (Given)

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

  2. Equation (By Corresponding Parts of Congruent Triangles)Therefore, Equation as alternate interior angles are equal.Now, Equation (co-interiors angles) Equation Equation

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

    Therefore Equation (By Side-Angle-Side congruence condition)

  4. Equation Equation (M is mid-point) Equation

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