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In Fig. 7.21, AC = AE, AB = AD and BAD = EAC. Show that BC = DE. Q.6

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This is very important question of ncert class 9th of chapter triangles. How I solve the best solution of exercise 7.1 question number 6. Please help me to solve this in a easy and best way ad give me the simple method of this question. In Fig. 7.21, AC = AE, AB = AD and BAD = EAC. Show that BC = DE.

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  1. Solution:

    It is given in the question that AB = AD, AC = AE, and ∠BAD = ∠EAC

    To prove:

    The line segment BC and DE are similar i.e. BC = DE

    Proof:

    We know that BAD = EAC

    Now, by adding DAC on both sides we get,

    BAD + DAC = EAC +DAC

    This implies, BAC = EAD

    Now, ΔABC and ΔADE are similar by SAS congruency since:

    (i) AC = AE (As given in the question)

    (ii) BAC = EAD

    (iii) AB = AD (It is also given in the question)

    ∴ Triangles ABC and ADE are similar i.e. ΔABC ΔADE.

    So, by the rule of CPCT, it can be said that BC = DE.

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