Chapter 7. Triangles
Exercise 7.1
Q1. In quadrilateral ACBD, AC = AD and AB bisects ∠A (see Fig. 7.16). Show that Δ ABC ≅ Δ ABD.
Solution:
Given: AC = AD and AB bisects ∠A
To prove: Δ ABC ≅ Δ ABD.
Proof: In Δ ABC and Δ ABD.
AC = AD [given]
∠CAB = ∠BAD [AB bisect ∠A]
AB = AB [Common]
By SAS Congruence Criterion Rule
Δ ABC ≅ Δ ABD
BC = BD [By CPCT] Proved
Q2. ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig.7.17). Prove that
(i) Δ ABD ≅ Δ BAC
(ii) BD = AC
(iii) ∠ ABD = ∠ BAC
Solution:
Given: ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA
To prove:
(i) Δ ABD ≅ Δ BAC
(ii) BD = AC
(iii) ∠ ABD = ∠ BAC
Proof: (i) In Δ ABD and Δ BAC
AD = BC [given]
∠ DAB = ∠ CBA [given]
AB = AB [Common]
By SAS Congruency Criterion Rule
Δ ABD ≅ Δ BAC
(ii) BD = AC [CPCT]
(iii) ∠ ABD = ∠ BAC [CPCT]
Q3. AD and BC are equal perpendiculars to a line segment AB (see the given figure). Show that CD bisects AB.
Solution:
Given: AD and BC are equal perpendiculars to a line segment AB.
To prove: CD bisects AB.
Proof:
In ∆BOC and ∆AOD
∠ BOC = ∠AOD (Vertically opposite angles)
∠CBO = ∠DAO (Each 90º)
BC = AD (Given)
By AAS Congruence Criterion Rule
∆BOC ≅ ∆AOD
BO = AO (By CPCT)
Hence, CD bisects AB.
Q4. l and m are two parallel lines intersected by another pair of parallel lines p and q (See the given figure). Show that ∆ABC ≅ ∆CDA
Solution:
Given: l and m are two parallel lines intersected by another pair of parallel lines p and q.
To prove: ∆ABC ≅ ∆CDA
Proof:
In ∆ABC and ∆CDA,
∠ BAC = ∠DCA (Alternate interior angles, as p || q)
AC = CA (Common)
∠ BCA = ∠DAC (Alternate interior angles, as l || m)
By AAS Congruence Criterion Rule
∆ABC ≅ ∆CDA
Q5. Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of a (see the given figure). Show that: (i) ∆APB ≅ ∆AQB (ii) BP = BQ or B is equidistant from the arms of ∠A.
Solution:
Given: Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of a.
To prove:
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
Proof:
In ∆APB and ∆AQB,
∠ APB = ∠AQB (Each 90º)
∠ PAB = ∠QAB (l is the angle bisector of A)
AB = AB (Common)
By AAS Congruence Criterion Rule
∆APB ≅ ∆AQB
BP = BQ [CPCT]
it can be said that B is equidistant from the A.
Q6. In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Solution:
Given: AC = AE, AB = AD and ∠BAD = ∠EAC.
To prove: BC = DE.
Proof: ∠BAD = ∠EAC
BAD + DAC = EAC + DAC
BAC = DAE
In ∆BAC and ∆DAE
AC = AE (Given)
AB = AD (Given)
∠BAC = ∠DAE (proved above)
By SAS Congruence Criterion Rule
∆BAC ≅ ∆DAE
BC = DE (CPCT)
Q7. AB 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 the given figure).
Show that:
(i) ∆DAP ≅ ∆EBP
(ii) AD = BE
Solution:
Given: AB 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.
To prove:
(i) ∆DAP ≅ ∆EBP
(ii) AD = BE
Proof: In ∆ DPA and ∆ EPB
∠EPA = ∠DPB
EPA + DPE = DPB + DPE
∠ DPA = ∠EPB
∠BAD =∠ ABE (Given)
∠EPA = ∠DPB (Given)
AP =BP (P is the midpoint of AB)
By AAS Congruence Criterion Rule
∆DAP ≅ ∆EBP
AD = BE (CPCT)