Chapter 7 Triangles
Ex 7.1 Class 9 Maths Question
1.
In quadrilateral ACBD, AC = AD and AB
bisects ∠ A (see figure). Show that ∆ABC ≅ ∆ABD. What can you
say about BC and BD?
Solution:
In quadrilateral ACBD, we have AC = AD and AB being the bisector of ∠A.
Now, In ∆ABC and ∆ABD,
AC = AD (Given)
∠ CAB = ∠ DAB ( AB
bisects ∠ CAB)
and AB = AB (Common)
∴ ∆ ABC ≅ ∆ABD (By
SAS congruence axiom)
∴ BC = BD (By
CPCT)
Ex 7.1 Class 9 Maths Question 2.
ABCD is a quadrilateral in which AD = BC
and ∠ DAB
= ∠ CBA
(see figure). Prove that
(i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD =
∠ BAC
Solution:
In quadrilateral ACBD, we have AD = BC and ∠ DAB = ∠ CBA
(i) In
∆ ABC and ∆ BAC,
AD = BC (Given)
∠DAB = ∠CBA (Given)
AB = AB (Common)
∴ ∆ ABD ≅ ∆BAC (By
SAS congruence)
(ii)
Since ∆ABD ≅ ∆BAC
⇒ BD = AC [By
C.P.C.T.]
(iii)
Since ∆ABD ≅ ∆BAC
⇒ ∠ABD = ∠BAC [By
C.P.C.T.]
Ex 7.1 Class 9 Maths Question
3.
AD and BC are equal perpendiculars to a
line segment AB (see figure). Show that CD bisects AB.
Solution:
In ∆BOC and ∆AOD, we have
∠BOC = ∠AOD
BC = AD [Given]
∠BOC = ∠AOD
[Vertically opposite angles]
∴ ∆OBC ≅ ∆OAD [By
AAS congruency]
⇒ OB = OA [By
C.P.C.T.]
i.e., O is the mid-point of AB.
Thus, CD bisects AB.
Ex 7.1 Class 9 Maths Question
4.
l and m are two parallel lines intersected
by another pair of parallel lines p and q (see figure). Show that ∆ABC = ∆CDA.
Solution:
∵ p || q and
AC is a transversal,
∴ ∠BAC = ∠DCA …(1)
[Alternate interior angles]
Also l || m and AC is a transversal,
∴ ∠BCA = ∠DAC …(2)
[Alternate interior angles]
Now, in ∆ABC and ∆CDA, we have
∠BAC = ∠DCA [From
(1)]
CA = AC [Common]
∠BCA = ∠DAC [From
(2)]
∴ ∆ABC ≅ ∆CDA [By
ASA congruency]
Ex 7.1 Class 9 Maths Question
5.
Line l is the bisector of an ∠ A and ∠ B is any point on
l. BP and BQ are perpendiculars from B to the arms of LA (see figure). Show
that
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms ot ∠A.
Solution:
We have, l is the bisector of ∠QAP.
∴ ∠QAB = ∠PAB
∠Q = ∠P [Each 90°]
∠ABQ = ∠ABP
[By angle sum property of A]
Now, in ∆APB and ∆AQB, we have
∠ABP = ∠ABQ [Proved
above]
AB = BA [Common]
∠PAB = ∠QAB [Given]
∴ ∆APB ≅ ∆AQB [By
ASA congruency]
Since ∆APB ≅ ∆AQB
⇒ BP = BQ [By
C.P.C.T.]
i. e., [Perpendicular distance of B from AP]
= [Perpendicular distance of B from AQ]
Thus, the point B is equidistant from the arms of ∠A.
Ex 7.1 Class 9 Maths Question
6.
In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC =
DE.
Solution:
We have, ∠BAD = ∠EAC
Adding ∠DAC on both sides, we have
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠DAE
Now, in ∆ABC and ∆ADE. we have
∠BAC = ∠DAE [Proved
above]
AB = AD [Given]
AC = AE [Given]
∴ ∆ABC ≅ ∆ADE [By
SAS congruency]
⇒ BC = DE [By
C.P.C.T.]
Ex 7.1 Class 9 Maths Question
7.
AS is a line segment and P is its
mid-point. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB. (see figure).
Show that
(i) ∆DAP ≅ ∆EBP
(ii) AD = BE
Solution:
We have, P is the mid-point of AB.
∴ AP = BP
∠EPA = ∠DPB [Given]
Adding ∠EPD on both sides, we get
∠EPA + ∠EPD = ∠DPB + ∠EPD
⇒ ∠APD = ∠BPE
(i)
Now, in ∆DAP and ∆EBP, we have
∠PAD = ∠PBE [ ∵∠BAD = ∠ABE]
AP = BP [Proved above]
∠DPA = ∠EPB [Proved
above]
∴ ∆DAP ≅ ∆EBP [By
ASA congruency]
(ii)
Since, ∆ DAP ≅ ∆ EBP
⇒ AD = BE [By
C.P.C.T.]
Ex 7.1 Class 9 Maths Question
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 figure). Show that
(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC
is a right angle
(iii) ∆DBC ≅ ∆ACB
(iv) CM =
Solution:
Since M is the mid – point of AB.
∴ BM = AM
(i) In
∆AMC and ∆BMD, we have
CM = DM [Given]
∠AMC = ∠BMD
[Vertically opposite angles]
AM = BM [Proved above]
∴ ∆AMC ≅ ∆BMD [By
SAS congruency]
(ii)
Since ∆AMC ≅ ∆BMD
⇒ ∠MAC = ∠MBD [By
C.P.C.T.]
But they form a pair of alternate interior angles.
∴ AC || DB
Now, BC is a transversal which intersects parallel lines AC and DB,
∴ ∠BCA + ∠DBC = 180°
[Co-interior angles]
But ∠BCA = 90°
[∆ABC is right angled at C]
∴ 90° + ∠DBC = 180°
⇒ ∠DBC = 90°
(iii)
Again, ∆AMC ≅ ∆BMD [Proved above]
∴ AC = BD [By
C.P.C.T.]
Now, in ∆DBC and ∆ACB, we have
BD = CA [Proved above]
∠DBC = ∠ACB [Each
90°]
BC = CB [Common]
∴ ∆DBC ≅ ∆ACB [By
SAS congruency]
(iv)
As ∆DBC ≅ ∆ACB
DC = AB [By C.P.C.T.]
But DM = CM [Given]
∴ CM =
⇒ CM =
