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 BAD=ABE and EPA=DPB (see Fig). Show that

  1. ΔDAPΔEBP

  2. AD=BE

Give 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

Give Ab Is a Line Segment and P Is Its Mid-Point.

Give 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. BAD=ABEandEPA=DPB

  1. EPA=DPB (Adding DPE both sides) EPA+DPE=DPB+DPE DPA=EPB

    In ΔDAPΔEBP , DPA=EPB AP=BP (P is mid-point of AB) BAD=ABE (Given)Therefore, ΔDAPΔEBP (By Angle-Side-Angle congruence condition.

  2. AD=BE (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

Give 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.

Give Right Triangle ABC

Give 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

C=90°, M is the mid-point of AB and DM=CM

  1. In ΔAMC and ΔBMD, AM=BM (M is the mid-point) CMA=DMB (Vertically opposite angles) CM=DM (Given)

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

  2. ACM=BDM (By Corresponding Parts of Congruent Triangles)Therefore, ACBD as alternate interior angles are equal.Now, ACB+DBC=180° (co-interiors angles) 90°+B=180° DBC=90°

  3. In ΔDBC and ΔACB, BC=CB (Common) ACB=DBC (Right angles) DB=AC (By Corresponding Parts of Congruent Triangles, already proved)

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

  4. DC=AB(ΔDBCΔACB) DM=CM=AM=BM (M is mid-point) DM+CM=AM+BMCM+CM=ABCM=12AB

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