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1. Chapter 7 Class 9 Triangles
2. Serial order wise
3. Ex 7.1

Transcript

Ex7.1, 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 the given figure). Show that: ΔAMC ≅ ΔBMD Given: ∠ ACB = 90° M is the mid-point of AB So, AM = BM Also, DM = CM To prove: ΔAMC ≅ ΔBMD Proof: Lines CD & AB intersect… So, ∠AMC = ∠BMD In ΔAMC and ΔBMD, AM = BM ∠AMC = ∠BMD CM = DM ∴ ΔAMC ≅ ΔBMD Ex7.1, 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 the given figure). Show that: (ii) ∠DBC is a right angle. From part 1, ΔAMC ≅ ΔBMD ∴ ∠ACM = ∠BDM But ∠ACM and ∠BDM are alternate interior angles for lines AC & BD If a transversal intersects two lines such that pair of alternate interior angles is equal, then lines are parallel. So, BD || AC Now, Since DB || AC and Considering BC as transversal , ⇒ ∠DBC + ∠ACB = 180° ⇒ ∠DBC + 90° = 180° ⇒ ∠DBC = 180° – 90° ⇒ ∠DBC = 90° Hence, ∠ DBC is a right angle Hence proved Ex 7.1, 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 the given figure). Show that: (iii) ΔDBC ≅ ΔACB From part 1, ΔAMC ≅ ΔBMD AC = BD In ΔDBC and ΔACB, DB = AC ∠DBC = ∠ACB BC = CB ∴ ΔDBC ≅ ΔACB Ex7.1, 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 the given figure). Show that: (iv) CM = 1/2 AB From part 3, ΔDBC ≅ ΔACB ∴ DC = AB 1/2 DC = 1/2 AB CM = 1/2 AB Hence proved

Ex 7.1

Chapter 7 Class 9 Triangles
Serial order wise 