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Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR. Q.14

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What is the best method to solve the problem of Exercise 6.3 of triangles, how to solve this problem in easy way Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.

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  1. Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that;

    AB/PQ = AC/PR = AD/PM

    We have to prove, ΔABC ~ ΔPQR

    Let us construct first: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.

    Ncert solutions class 10 chapter 6-25

    In ΔABD and ΔCDE, we have

    AD = DE  [By Construction.]

    BD = DC [Since, AP is the median]

    and, ∠ADB = ∠CDE [Vertically opposite angles]

    ∴ ΔABD ≅ ΔCDE [SAS criterion of congruence]

    ⇒ AB = CE [By CPCT] …………………………..(i)

    Also, in ΔPQM and ΔMNR,

    PM = MN [By Construction.]

    QM = MR [Since, PM is the median]

    and, ∠PMQ = ∠NMR [Vertically opposite angles]

    ∴ ΔPQM = ΔMNR [SAS criterion of congruence]

    ⇒ PQ = RN [CPCT] ………………………………(ii)

    Now, AB/PQ = AC/PR = AD/PM

    From equation (i) and (ii),

    ⇒CE/RN = AC/PR = AD/PM

    ⇒ CE/RN = AC/PR = 2AD/2PM

    ⇒ CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN]

    ∴ ΔACE ~ ΔPRN [SSS similarity criterion]

    Therefore, ∠2 = ∠4

    Similarly, ∠1 = ∠3

    ∴ ∠1 + ∠2 = ∠3 + ∠4

    ⇒ ∠A = ∠P …………………………………………….(iii)

    Now, in ΔABC and ΔPQR, we have

    AB/PQ = AC/PR (Already given)

    From equation (iii),

    ∠A = ∠P

    ∴ ΔABC ~ ΔPQR [ SAS similarity criterion]

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