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Class Nine NCERT Solutions Maths Chapter 7

Chapter 7: Triangles (NCERT Solutions)

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 :

(i) Δ AMC ≅ Δ BMD

(ii) ∠ DBC is a right angle.

(iii) Δ DBC ≅ Δ ACB

(iv) CM = 1/2 AB

Sol. In ΔBMD and ΔAMC, we have

(i) DM = CM [Given]

BM = AM [Since M is the mid-point of AB]

∠DMB = ∠AMC [Vertically opposite angles]

∴ ΔAMC ≅ ΔBMD (By SAS congruence) Proved.

(ii) AC || BD [Since, ∠DBM and ∠CAM are alternate angles]

⇒ ∠DBC + ∠ACB = 180° [Sum of co-interior angles)]

⇒ ∠DBC + 90° = 180° [Science ∠ACB = 90°]

⇒ ∠DBC = 90° Proved.

(iii) In ΔDBC and ΔACB, we have

DB = AC [CPCT]

BC = BC [Common]

∠DBC = ∠ACB [Each = 90°]

∴ ΔDBC ≅ ΔACB [By SAS congruence] Proved.

(iv) ∴ AB = CD [CPCT]

⇒ 12 AB = 1/2 CD

Hence, 1/2 AB = CM

[Science CM = 1/2 CD] Proved.

TEXTBOOK’S EXERCISE – 7.2

1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠ B and ∠C intersect each other at O. Join A to O. Show that :

(i) OB = OC (ii) AO bisects ∠ A.

Sol. (i) AC = AB ⇒ ∠ABC = ∠ACB [Angles opposite to equal sides are equal]

⇒ 12 ∠ABC = 12 ∠ACB

⇒ ∠CBO = ∠BCO [Since, OB and OC are bisectors of ∠B and ∠C respectively]

⇒ OB = OC [Sides opposite to equal angles are equal]

Again, 12 ∠ABC = 12 ∠ACB

⇒ ∠ABO = ∠ACO [∴ OB and OC are bisectors of ∠B and ∠C respectively]

In ΔABO and ΔACO, we have

AB = AC [Given]

∠ABO = ∠ACO [Proved above]

OB = OC [Proved above]

∴ ΔABO ≅ ΔACO [SAS congruence]

⇒ ∠BAO = ∠CAO [CPCT]

⇒ AO bisects ∠A Proved.

2. In Δ ABC, AD is the perpendicular bisector of BC (see Fig.). Show that Δ ABC is an isosceles triangle in which AB = AC.

Sol. In ΔABD and ΔACD, we have

BD = CD [since AD bisects BC]

∠ADB = ∠ADC [Each = 90°]

AD = AD [Common]

∴ ΔABD ≅ ΔACD [SAS]

∴ AB = AC [CPCT]

Hence, ΔABC is an isosceles triangle. Proved.

3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig.). Show that these altitudes are equal.

Sol. In ΔABC,

AC = AB [Given]

⇒ ∠B = ∠C [Angles opposite to equal sides are equal]

Now, in right triangles BFC and CEB,

∠BFC = ∠CEB [Each = 90°]

∠FBC = ∠ECB [Proved above]

BC = BC [Common]

∴ ΔBFC ≅ ΔCEB [AAS]

Hence, BE = CF [CPCT] Proved.

4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig.). Show that

(i) Δ ABE ≅ Δ ACF

(ii) AB = AC, i.e., ABC is an isosceles triangle.

Sol. (i) In ΔABE and ΔACF, we have

BE = CF [Given]

∠BAE = ∠CAF [Common]

∠BEA = ∠CFA [Each = 90°]

So, ΔABE ≅ ΔACF [AAS] Proved.

(ii) Also, AB = AC [CPCT]

i.e., ABC is an isosceles triangle Proved.

5. ABC and DBC are two isosceles triangles on the same base BC (see Fig.). Show that ∠ ABD = ∠ ACD.

Sol. In isosceles ΔABC, we have

AC = AB

∠ABC = ∠ACB …(i) [Angles opposite to equal sides are equal]

Now, in isosceles ΔDCB, we have

CD = BD

∠DBC = ∠DCB …(ii) [Angles opposite to equal sides are equal]

Adding (i) and (ii), we have

∠ABC + ∠DBC = ∠ACB + ∠DCB

⇒ ∠ABD = ∠ACD Proved.

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