NCERT Solution for Triangles 7.1

NCERT Solutions

Exercise 7.1 (Page 118)

1.  In quadrilateral ACBD, AC = AD and AB bisects ∠A )see figure). Show that ΔABC ≌ ΔABD. What can you say about BC and BD?

Ans. In quadrilateral ABCD we have

                  AC = AD

            and AB being the bisector of ∠A.

            Now, in ΔABC and ΔABD,

                  AC = AD

[Given]

                  AB = AB

[Common]

∠CAB = ∠DAB [∴ AB bisects ∠CAD]

            ∴ Using SAS criteria, we have

                  ΔABC ≌ ΔABD.

            ∴ Corresponding parts of congruent triangles (c.p.c.t) are equal.

            ∴ BC = BD.

2.  ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Figure). Prove that

      (i) ΔABD ≌ ΔBAC

      (ii) BD = AC

      (iii) ∠ABD = ∠BAC.

Ans. (i) In quadrilateral ABCD, we have AD = BC and

                ∠DAB = ∠CBA.

                In ΔABD and ΔBAC,

                AD = BC

[Given]

                AB = BA

[Common]

                ∠DAB = ∠CBA

[Given]

                ∴ Using SAS criteria, we have ΔABD ≌ ΔBAC

          (ii) ∵ ΔABD ≌ ΔBAC

                ∴ Their corresponding parts are equal.

                ⇒ BD = AC

          (ii) Since ΔABD ≌ ΔBAC

                ∴ Their corresponding parts are equal.

                ⇒ ∠ABD = ∠BAC.

3.  AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.

Ans. We have ∠ABC = 90° and ∠BAD = 90°

          Also AB and CD intersect at O.

          ∴ Vertically opposite angles are equal.

          Now, in ΔOBC and ΔOAD, we have

                  ∠ABC = ∠BAD

[each = 90°]

                  BC = AD

[Given]

                  ∠BOC = ∠AOD

[vertically opposite angles]

          ∴ Using ASA criteria, we have

                  ΔOBC ≌ ΔOAD

          ⇒ OB = OA

[c.p.c.t]

          i.e. O is the mid-point of AB

          Thus, CD bisects AB.

4.  l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ΔABC ≌ ΔCDA.

Ans. ∵l || m and AC is a transversal.

          ∴ ∠BAC = ∠DCA

[Alternate interior angles]

          Also p || q and AC is a transversal,

          ∴ ∠BCA = ∠DAC

[Alternate interior angles]

          Now, in ΔABC and ΔCDA,

                ∠BAC = ∠DCA

[Proved]

                ∠BCA = ∠DAC

[Proved]

                CA = AC

[Common]

          ∴ Using ASA criteria, we have ΔABC ≌ ΔCDA

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 Figure). Show that:

          (i) ΔAPB ≌ ΔAQB

          (ii) BO = BQ or B is equidistant from the arms of ∠A.

Ans. We have, l as the bisector of QAP.

          ∴ ∠QAB = ∠PAB

          ∴ ∠Q = ∠P

[each = 90°]

          ⇒ Third ∠ABQ = Third ∠ABP

          (i) Now, in ΔAPB and ΔAQB, we have

                AB = AB

[Common]

                ∠ABP = ∠ABQ

[Proved]

                ∠PAB = ∠QAB

[Proved]

                ∴ Using SAS criteria, we have

                ΔAPB ≌ ΔAQB

          (ii) Since ΔAPB ≌ ΔAQB

                ∴ Their corresponding angles are equal.

                ⇒ BP = BQ

                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.

6.  In the figure, AC = AE, AB = AD and ∠BAD = ∠ESC. Show that BC = DE.

Ans. We have ∠BAD = ∠EAC

            Adding ∠DAC on both sides, we have

            ∠BAD + ∠DAC = ∠EAC + ∠DAC

            ⇒ ∠BAC = ∠DAE

            Now, in ΔABC and ΔADE, we have

                ∠BAC = ∠DAE

[Proved]

                AB = AD

[Given]

                AC = AE

[Given]

            ∴ ΔABC ≌ ΔADE

[Using SAS criteria]

            Since ΔABC ≌ ΔADE, therefore, their corresponding parts are equal.

            ⇒ BC = DE.

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 figure). Show that

      (i) ΔDAP ≌ ΔEBP

      (ii) AD = BE

Ans. 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 ΔDAP ≌ ΔEBP, we have

                  AP = BP

[Proved]

            ∠PAD = ∠ PBE

[∴ It is given that ∠BAD = ∠ABE]

            ∠DPA = ∠EPB

[Proved]

            ∴ Using ASA criteria, we have

            ΔDAP ≌ ΔEBP

            (ii) Since,

                 ∴ Their corresponding parts are equal.

                 ⇒ AD = BE.

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) ΔAMC ≌ ΔBMD

      (ii) ∠DBC is a right angle.

      (ii) ΔDBC ≌ ΔACB

      (iv) CM =  AB

Ans. ∵ M is the mid-point of AB.

            ∴ BM = AM

[Given]

            (i) In ΔAMC and ΔBMD, we have

                CM = DM

[Given]

                AM = BM

[Proved]

                ∠AMC = ∠BMD

[Vertically opposite angles]

                ∴ ΔAMC ≌ ΔBMD

[SAS criteria]

            (ii) ∴ΔAMC ≌ ΔBMD

                ∴ Their corresponding parts are equal.

                ⇒ ∠MAC = ∠MBD

                But they form a pair of alternate interior angles.

                ∴ AC || DB

                Now, BC is a transversal which intersecting parallel lines AC and DB,

                ∴ ∠BCA + ∠DBC = 180°

                But ∠BCA = 90°

[∴ ΔABC is right angled at C]

                ∴ 90° + ∠DBC = 180°

                or ∠DBC = 180° – 90° = 90°

                Thus, ∠DBC = 90°

            (iii) Again, ΔAMC ≌ ΔBMD

[Proved]

                ∴ AC = BD

[c.p.c.t]

                Now, in ΔDBC and ΔACB, we have

                ∠DBC = ∠ACB

[Each = 90°]

                BD = CA

[Proved]

                BC = CB

[Common]

                ∴ Using SAS criteria, we have

                            ΔDBC ≌ ΔACB

            (iv) ∴ ΔDBC ≌ ΔACB

                ∴ Their corresponding parts are equal.

                ⇒ DC = AB

                But DM = CM

[Given]