Triangles (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,
∠ 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,
∠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°]
∠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.

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:

Adding ∠DAC on both sides, we have
∠BAD + ∠DAC = ∠EAC + ∠DAC
Now, in ∆BAC and ∆DAE.
∠BAC = ∠DAE          [Proved above]
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 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 = ½DC = ½AB
⇒ CM = ½AB.