MathematicsState true or false:In the ΔABC, P and Q are points on AB and AC such that PQ || BC, then the median AD bisects PQ.

State true or false:


In the ΔABC, P and Q are points on AB and AC such that PQ || BC, then the median AD bisects PQ.


  1. A
    True
  2. B
    False 

    Fill Out the Form for Expert Academic Guidance!l



    +91



    Live ClassesBooksTest SeriesSelf Learning



    Verify OTP Code (required)

    I agree to the terms and conditions and privacy policy.

    Solution:

    It is given that in the ΔABC, P and Q are points on AB and AC such that PQ || BC.
    The required figure is,
    By AA similarity criterion if two triangles have 2 pairs of congruent angles, then the triangles are similar.
    By corresponding angle axiom if a transversal intersects two parallel lines, then each pair of corresponding angles are equal.
    As PQ || BC and PE || BD so,
    ∠APE=∠ABD   …………(1)
    ∠AEP=∠ADB   …………(2)
    From equation (1) and (2),
    ΔAPE ∼ ΔABD  (By AA similarity criterion)
    We know that the corresponding sides of similar triangles have the same ratios, then
    ⇒PEBD=AEAD   ………..(3)
    As PQ || BC and EQ || DC so,
    ∠AQE=∠ACD   …………(4)
    ∠AEQ=∠ADQ   …………(5)
    From equation (4) and (5),
    ΔAEQ ∼ ΔADC  (By AA similarity criterion)
    ⇒EQDC=AEAD   ………..(6)
    Combine equations (3) and (6),
    PEBD=EQDC
    But BD=DC as AD is the median.
    ⇒PEDC=EQDC
    ⇒PE=QE
    ⇒ The median AD bisects PQ.
    Therefore, it is true that in the ΔABC, P and Q are points on AB and AC such that PQ || BC, then the median AD bisects PQ.
    Hence, option 1 is correct.
     
    Chat on WhatsApp Call Infinity Learn