NCERT Solution for Triangles 7.3

Exercise 7.3 (Page 128)
1.  ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that
      (i) ΔABD ≌ ΔACD
      (ii) ΔABP ≌ ΔACP
      (iii) AP bisects ∠A as well as ∠D.
      (iv) AP is the perpendicular bisector of BC.
Ans. (i) In ΔABD and ΔACD, we have
                AB = AC
                AD = AD
                BD = CD
                ΔABD ≌ ΔACD
            (ii) In ΔABP and ΔACP, we have
                AB = AC
[Given]
                ∴ AB = AC ⇒ ∠B = ∠C
[∴ Angle opposite to equal sides are equal]
                AP = AP
[Common]
                ∴ ΔABP ≌ ACP
[SAS Criteria]
            (iii) Since, ΔABP ≌ ΔACP
                ∴ Their corresponding parts are congruent.
                ⇒ ∠BAP = ∠CAP
                ∴ AP is the bisector of ∠A.
...(1)
                Again, in ΔBDP and ΔCDP, we have
                BD = CD
[Given]
                ∠DBP = ∠CDP
[Angles opposite to equal sides]
                DP = DP
[Common]
                ⇒ ∠BDP ≌ ∠CDP
                ∴ ∠BDP ≌ ∠CDP [c.p.c.t]
                ⇒ DP (or AP) is the bisector of ∠D.
...(2)
                From (1) and (2), Ap is the bisector of ∠A as well as ∠D.
            (iv) ∵ΔABP ≌ ΔACP
                ∴ Their corresponding parts are equal.
                ⇒∠APB = ∠APC
But ∠APB + ∠APC = 180° [Linear pair]
                ∴ ∠APB = ∠APC = 90°
                ⇒ AP ⊥ BC
                ⇒ AP is the perpendicular bisector of BC
2.  AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
      (i) AD bisects BC                        (ii) AD bisects ∠A
Ans. (i) In ΔABD and ΔACD, we have
                AB = AC
[Given]
                ∠B = ∠C
[Angles opposite to equal sides]
                AD = AD
[Common]
                ∴ ΔABD ≌ ΔACD
                ⇒ Their corresponding parts are equal.
                ∴ BD = CD
                ⇒ D is the mid-point of BC or AD bisects BC.
          (ii) Since, ΔABD ≌ ΔACD,
                ∴ Their corresponding parts are congruent.
                ⇒ ∠BAD = ∠CAD
                ⇒ AD bisects ∠A.
3.  Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see figure). Show that
      (i) ΔABM ≌ ΔPQN                        (ii) ΔABC ≌ ΔPQR
Ans. In ΔABC, AM is a median
[Given]
            
...(1)
            In ΔPQR, PN is a median.
            
...(2)
            ∴ BC = QR
[Given]
            
            ⇒BM =QN
[From (1) and (2)]
          (i) In ΔABM and ΔPQN, we have
                ∴AB = PQ
[Given]
                AM = PN
[Given]
                BM = QN
[Proved]
                ∴ ΔABM ≌ ΔPQN
[SSS criteria]
          (ii) ∵ΔABM ≌ ΔPQN
                ∴ Their corresponding parts are congruent.
                ⇒ ∠B = ∠Q
                Now, in ΔABC and ΔPQR, we have
                ∠B = ∠Q
[Proved]
                AB = PQ
[Given]
                BC = QR
[Given]
                ΔABC ≌ ΔPQR
4.  BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Ans. BE ⊥ AC
[Given]
          ∴ ΔBEC is a right triangle such that
          ∠BEC = 90°
          Similarly, ∠CFB = 90°
          Now, in right ΔBEC and right ΔCFB, we have
          BE = CF
[Given]
          BC = CB
[Common]
          ∴ Using RHS criteria, ΔBEC ≌ ΔCFB
          ∴ Their corresponding parts are equal.
          ⇒ ∠BCE = ∠CBF
          or ∠BCA = ∠CBA
          Now, in ΔABC, ∠BCA = ∠CBA
          ∴ Their opposite sides are equal.
          ⇒ AB = AC
          ∴ ΔABC is an isosceles triangle.
5.  ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Ans. We have AP ⊥ BC
[Given]
          ∴ ∠APB = 90° and APC = 90°
          In ΔABP and ΔACP, we have
          ∠APB = ∠APC
[each = 90°]
          AB = AC
[Given]
          AP = AP
[Common]
          ∴ Using RHS criteria,ΔABP ≌ ΔACP
          ∴ Their corresponding parts are congruent.
          ⇒ ∠B = ∠C

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