Triangles
Chapter 7
Exercise 7.1
EX 7.1 QUESTION 1.
In quadrilateral ACBD, AC = AD and AB bisect A (see Fig. 7.16). Show that ΔABC ΔABD. What can you say about BC and BD?
Solution:
In ∆ABC and ∆ABD,
AC = AD [Given]
∠ CAB = ∠ DAB [AB bisects ∠A]
and AB = AB [Common]
∴ ∆ ABC ≅ ∆ABD [SAS congruence rule]
∴ BC = BD [By CPCT]
EX 7.1 QUESTION 2.
ABCD is a quadrilateral in which AD = BC and DAB = CBA (see Fig. 7.17). Prove that
(i) ΔABD ΔBAC
(ii) BD = AC
(iii) ABD = BAC.
Solution:
(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 QUESTION 3.
AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.
Solution:
In ∆OCB and ∆ODA, we have
∠CBO = ∠DAO [Each 90°]
BC = AD [Given]
∠BOC = ∠AOD [Vertically opposite angles]
∴ ∆OCB ≅ ∆ODA [By AAS congruency]
⇒ BO = AO [By C.P.C.T.]
CD bisects AB.
EX 7.1 QUESTION 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ΔABC ΔCDA.
Solution:
∠BAC = ∠ACD [Alternate angles]
CA = AC [Common]
∠BCA = ∠DAC [Alternate angles]
∴ ∆ABC ≅ ∆CDA [By ASA congruency rule]
EX 7.1 QUESTION 5.
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 Fig. 7.20). Show that:
(i) ΔAPB ΔAQB
(ii) BP = BQ or B is equidistant from the arms of A.
Solution:
In ∆APB and ∆AQB
∠ABP = ∠AQB [Proved above]
AB = BA [Common]
∠PAB = ∠QAB [Given]
∴ ∆APB ≅ ∆AQB [By ASA congruency]
(ii) BP = BQ [By C.P.C.T.]
EX 7.1 QUESTION 6.
In Fig. 7.21, AC = AE, AB = AD and BAD = EAC. Show that BC = DE.
Solution:
∠BAD = ∠EAC [Given]
Adding ∠DAC on both sides, we have
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠EAD
Now, in ∆BAC and ∆DAE.
∠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 QUESTION 7.
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 Fig. 7.22). Show that
(i) ΔDAP ΔEBP
(ii) AD = BE
Solutions:
(i)∠EPA = ∠DPB [Given]
Adding ∠EPD on both sides, we get
∠EPA + ∠EPD = ∠DPB + ∠EPD
⇒ ∠APD = ∠BPE
(i)In ∆DAP and ∆EBP,
∠A = ∠B [Given]
AP = BP [P is the mid-point of AB.]
∠APD = ∠BPE [Proved above]
∴ ∆DAP ≅ ∆EBP [By ASA congruency]
(ii) AD = BE [By C.P.C.T.]
EX 7.1 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 Fig. 7.23). Show that:
(i) ΔAMC ΔBMD
(ii) DBC is a right angle.
(iii) ΔDBC ΔACB
(iv) CM = ½ AB
Solution:
M is the midpoint of AB.
∴ BM = AM
(i) In ∆AMC and ∆BMD,
CM = DM [Given]
∠AMC = ∠BMD [Vertically opposite angles]
AM = BM [Proved above]
∴ ∆AMC ≅ ∆BMD [By SAS congruency]
(ii) Since ∆AMC ≅ ∆BMD
⇒ ∠CAM = ∠DBM [By C.P.C.T.]
alternate interior angles (∠CAM = ∠DBM) are equal,
∴ AC || DB
Now, BC is a transversal which intersects parallel lines AC and DB,
∴ ∠ACB + ∠DBC = 180° [Co-interior angles]
∠ACB = 90° [∆ABC is right angled at C]
∴ 90° + ∠DBC = 180°
⇒ ∠DBC = 90°
(iii) ∆AMC ≅ ∆BMD [Proved above]
∴ AC = BD [By C.P.C.T.]
In ∆DBC and ∆ACB,
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 = ½
⇒ CM = ½
