NCERT Solutions for Class 9th Maths: Chapter 7 Triangles Exercise 7.3

Discussion in 'CBSE Class 9 Maths Help' started by saralas, May 18, 2014.

  1. saralas

    saralas Member

    [​IMG]ABC and [​IMG]DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show that
    (i) [​IMG]ABD [​IMG][​IMG]ACD
    (ii) [​IMG]ABP [​IMG][​IMG]ACP
    (iii) AP bisects [​IMG]A as well as [​IMG]D.
    (iv) AP is the perpendicular bisector of BC.

    NCERT Solutions for Class 9th Maths Exercise  7.3 Q1.JPG
  2. saralas

    saralas Member

    (i) In [​IMG]ABD and [​IMG]ACD
    AB = AC (given)
    BD = CD (given)
    AD = AD (common)
    [​IMG]

    (ii) In [​IMG]ABP and [​IMG]ACP
    AB = AC (given).
    [​IMG]BAP = [​IMG]CAP [from equation (1)]
    AP = AP (common)
    [​IMG]
  3. saralas

    saralas Member

    (iii) From equation (1)
    [​IMG]BAP = [​IMG]CAP
    Hence, AP bisect [​IMG]A
    Now in [​IMG]BDP and [​IMG]CDP
    BD = CD (given)
    DP = DP (common)
    BP = CP [from equation (2)]
    [​IMG]
    (iv) We have [​IMG]BDP [​IMG][​IMG]CDP
    [​IMG]

    Now, [​IMG]BPD + [​IMG]CPD = 180o (linear pair angles)

    [​IMG]BPD + [​IMG]BPD = 180o

    2[​IMG]BPD = 180o [from equation (4)]

    [​IMG]BPD = 90o ...(5)

    From equations (2) and (5), we can say that AP is perpendicular bisector of BC.

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