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]