Telangana SCERT Solution Class IX (9) Math Chapter 7 Triangles Exercise 7.1
1.)
Given,
AB = AD
AB bisects ∠A
∴ ∠BAC = ∠BAD
NOW,
FROM △ ABC and △ABD
i) Sides AC = AD
ii) ∠BAC = ∠BAD
iii) AB IS COMMON SIDE.
∴ △ABC ≅ △ABD by SAS
∴ BC = BC [∵△ABC ≅ △ABD]
2.)
Given,
AD = BC
∠DAB = ∠CBA
∴ In △ABC and △ABD
i) AD = BC (given)
ii) ∠DAB, ∠CBA (″)
iii) AB is common side.
∴ i) △ ABC ≅ △ABD by SAS.
ii) BD = AC [∵Corresponding side of congruent – triangles are equal]
(iii) ∠ABD = ∠ BAC [∵ Corresponding angels of congruent △ are equal]
3.)
Given,
AD > BC
AB ⊥ AD
AB ⊥ BC
∴ From the given information,
BC is parallel to AD Sine both are ⊥ to the same line segment –AB.
In △ ADO and △BCD
∴ i) ∠ADO and ∠BCD [alternate angels]
ii) ∠OAD = ∠ OBC = 90°
iii) BC = AD [given]
△ADO ≅ △ BCO by RHS.
∴ AO = DB [∵△ADO≅ △BCD]
Here we can conclude that – CD bisect – AB.
4.)
Given,
l ∥ m
P ∥ q
∴ From the given into,
In △ABC and △ACD
i) ∠ACB = ∠CAD [alternate angels]
ii) ∠ ACD = ∠ BAC [‘’ ‘’ ‘’]
iii) AC is common side.
∴ △ABC ≅ △ACD by ASA
5.)
Given,
AC = AE, AB = AD
∠BAD = ∠EAC
Now,
∠BAD + ∠DAC = ∠EAC+∠DAC [∵∠BAD = ∠ EAC
∴ ∠BAC = ∠EAD
∴ ∠In △ ABC and △ EAD
i) ∠BAC = ∠EAD
ii) AB = AD
iii) AE = AC
∴ ABC ≅ △EAD by SAS
∴ BC = DE since corresponding side of congruent – △s are equal.
6.)
Given,
∠ACB = 90°
M is midpoint of AB
∴ AM = BM
DM = CM
i) In △AMC and △BMD
i) DM = CM
ii) AM = BM
iii) ∠BMD = ∠AMC [vertically opposite angels]
∴ △AMC ≅ △BMD by SAS
ii) In △BDC and △ABC
i) ∠BDM= ∠CAM [∵△AMC ≅ △BMD]
ii) AC = BD [∵ ‘’ ‘’ ‘’]
iii) BC is common side…..
iii) ∴ △ABC ≅ △BDC by SAS
ii) ∴ ∠DBC = ∠ACB = 90° [∵△ABC ≅ △BDC]
∴ AB = DC [∵△ABC ≅ △BDC]
∴Now,
CM = ½ DC [∵DM = CM]
iv) CM = ½ AB [∵DC = AB]
7.)
Given,
ABCD is a square.
AB = BC = CD = AD
∴∠DAB = ∠ABC = ∠BCD = ∠CDA = 90°
△APB is an equilateral △
∴ AP = PB = AB
∠ABP = ∠BPA = ∠PAB =60°
In △APD and △BPC
i) AB = PB [△APB equilateral △]
ii) AD = BC [∴ ABCD is a Square]
iii) ∠DAP = ∠DAB – ∠PAB
= 90° – 60° = 30°
∠PBC = ∠ABC – ∠ABP
= 90° – 60°
= 30°
∴ ∠DAP = ∠PBC
∴ △APD ≅△BPC by SAS
8.)
Given,
DE = DF
D midpoint of BC
∴ BD = CD
DE ⊥ AB
∴ DE ⊥ BE
DF ⊥ AC
∴ DP ⊥ CF
∴ In △ BED and △CFD
i) DE = DF
ii) BD = CD
iii) ∠BED = ∠CFD = 90° [∵ DE ⊥ BE, DF ⊥CF]
∴ △ BED ≅ △CFD by RHS
9.)
From the given information,
Draw △ABC where
AD Bisects ∠A and side is C
Now,
Draw a parallel line from point – C parallel to AD
Extend BA so that it meets the parallel line from C at point E.
Now,
∠BAD = ∠DAC
BD = CD
∴∠BAD = ∠ABC [corresponding angel]
∠DAC = ∠ACE [alternate angels]
∴ AE = AC [∴ Sides opposite to equilateral of a △are equal]
Now,
In △ BCE, AD ∥ CE
D is the midpoint of BC
∴ A is the midpoint of BE by converse of midpoint theorem
∴ AB = AE
∴ AB = AC [∵ AE = AC]
∴ △ ABC is an isosceles triangle.
10.)
Given,
∠B right angle.
∠BCA = 2∠BAC.
Produce CB to D so that
CB = BD
∴ In △ABC and △ABD
i) BC = BD
ii) AB common side
iii) ∠ABC = ∠ABD = 90° [∴∠ABD is collinear angle to ∠ABC = 180° – 90° = 90°]
∴ △ABC ≅ △ABD by SAS.
Let,
∴ ∠CAB = ∠DAB = x [∴△ADC ≅ △ACD]
∴ ∠BCA = 2x
∴ ∠ CAD = ∠CAB +∠DAB = 2x
∴ ∠CAD = ∠BAC = 2x
∴ CD = AD [∴ Opposite sides of equal angels are equal]
AD = AC [∴△ABC ≅△ABD]
∴ CD = AC
∴ 2BC = AC [∵B midpoint – of CD]
Proved.