**Telangana SCERT Solution Class IX (9) Math Chapter 7 Triangles Exercise 7.3**

**1.)**

The adjacent figure represents △ABC of the question when △ABC is an isosceles triangle.

AD is the altitude to side BC.

∴ AB = AC

∠ADB =∠ADC = 90° [∵ AD altitude to side BC]

∴ In △ABD and △ ACD

i) AB = AC [given]

ii) AD is the common side

iii) ∠ADB = ∠ADC = 90°.

∴ △ABD ≅ △ADC by RHS

∴ BD = DC [∵ corresponding side of congruent triangles are equal]

i) ∴ AD bisects BC

∠BAD = ∠CAD [∵ Corresponding angles of congruent triangles are equal]

ii) ∴ AD bisects ∠A.

**2.)**

Given,

AB = PQ

AM = PN

BC = QR

AM and PN are medium of Side BC and QR respectively.

∴ BM = 1/2 BC

CN = 1/2 QR

∴ BM = CN [∵ BC = QR]

i) ∴ In △ABM and △PQN

i) AB = PQ [given]

ii) AM = PN [given]

iii) BM = CN

∴ △ABN ≅ △PQN by SSS

∴ ∠ABM or ∠ABC = ∠PQN or ∠PQR.

[∴ Corresponding angles of congruent △s are equal]

ii) ∴ In △ABC and △PQR

i) AB = PQ [given]

ii) ∠ABC = ∠PQR

iii) BGC = QR [given]

∴ △ABC ≅ △PQR by SAS

**3.)**

Given,

BE and CF altitudes of side AC and AB respectively and BE = CF

∴ ∠AEB = ∠AFC = 90° [∵BE and CF altitudes of △ABC]

Now,

In △ABE and △ACF by RHS congruency rule we can say that.

- i) Hypotenuse of △ABE and △ACF are equal.

∴ AB = AC

- ii) BE = CF given

iii) ∠AEB = ∠AFC = 90°

∴ In △ABC

AB = AC

∴ △ABC is an isosceles Triangle.

**4.)**

To construct an isosceles △with BC as base draw a line segment –BC

Now,

From point B and C draw an arc with sufficient – length that cuts cuts other at point A

∴ △ABC is an isosceles △ with AB = AC

Now,

Using a protector draw a perpendicular AP from base BC meeting point A

Now,

In △ABP and △APC

- i) AB = AC [∵△ABC is an isosceles triangle]
- ii) ∠APB = ∠APC = 90° [∵AP ⊥ to BC]

iii) AP is the common side.

∴ △ABP ≅ △APC by RHS

∴ ∠B = ∠C [∵ corresponding angles of congruent triangles are equal]

**5.)**

Given,

△ABC is an isosceles triangle.

∴ AB = AC

AB = AD = AC

In △ ABC

∠ABC = ∠ACB [Opposite angles of equal sides of △ are equal]

∴ ∠ABC + ∠ACB + ∠CAB = 180° [Sum of angles of △ is 180°]

Or, ∠ABC + ∠ACB + ∠CAB = 180° [∴∠ABC = ∠ACB] – (i)

In △ACD

∠ACD = ∠ADC [opposite angles of equal sides of a triangle are equal]

∠ACD + ∠ADC + ∠CAD = 180° [∵ sum of angles of △ is 180°]

∠ CAD = 180° – 2∠ACD [∵∠ACD = ∠ADC] – (ii)

Now,

∠CAB + ∠CAD = 180° [collinear angles]

Substituting the values of ∠CAB and ∠CAD from

Eqn (i) and (ii) we get

180° -2 ∠ACB + 180° – 2 ∠ACD = 180°

Or, – 2(∠ACB + ∠ACD) = 180° – 360°

Or, – 2 ∠BCD = – 180°

Or, ∠BCD = 90° Hence prove.

**6.)**

Draw a line segment AB used from point A draw another line segment AC ⊥AB using a protector such that AC = AB

Now,

Join BC we have △ABC as sight angled △with ∠A = 90°

∴ ∠B = ∠C [Opposite angles of equal side of a △ are equal]

Hence proved

**7.)**

In △ABC AB = BC = AC

∴ △ABC is equilateral △

∴ ∠A = ∠B [∵ AC =BC Opposite angles of equal side of a △ are equal]

∠B = ∠C [∵ AB = AC angles opposite of equal sides of a △ are equal]

∴ ∠A = ∠B = ∠C

Now,

∠A+∠B = ∠C = 180° [∵ Sum of angles of a △ is 180°]

Or, ∠A + ∠A + ∠A = 180

Or, 3∠A = 180° ∴∠A = ∠B = ∠C = 60°

Or, ∠A = 60°

Angles of equilateral triangles are each 60°.

**8.)**

Given,

△ABC is an isosceles △

Where,

AB = AC

AQ = AP

Now,

In △ PBA and △QCA

i) AP = AQ [given]

ii) AB = AC [given]

iii) ∠PAB = ∠QAC [Vertically opposite angles]

∴ △PBA ≅ △QCA by SAS

∴ PB = QC [∵ corresponding side of congruent triangles are equal]