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    In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see Figure). Show that: (i) Δ AMC ≅ Δ BMD (ii) ∠ DBC is a right angle. (iii) Δ DBC ≅ Δ ACB (iv) CM = (1/2) AB

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    sanjeev rao

    (i) In ΔAMC and ΔBMD, CM = DN [∵ Given] ∠AMC = ∠BMD [∵ Vertically Opposite Angles] AM = MB [∵ M is the mid-point of line segment AB] Hance, ΔAMC ≅ ΔBMD [∵ SAS Congruency Rule] (ii) ΔAMC ≅ ΔBMD [∵ Proved above] ∠CAM = ∠DMB [∵ Corresponding parts of congruent Triangles are equal ] Since, alternate angRead more

    (i) In ΔAMC and ΔBMD,
    CM = DN [∵ Given]
    ∠AMC = ∠BMD [∵ Vertically Opposite Angles]
    AM = MB [∵ M is the mid-point of line segment AB]
    Hance, ΔAMC ≅ ΔBMD [∵ SAS Congruency Rule]

    (ii) ΔAMC ≅ ΔBMD [∵ Proved above]
    ∠CAM = ∠DMB [∵ Corresponding parts of congruent Triangles are equal ]
    Since, alternate angles (∠CAM and ∠DMB) are equal, therefore AC ∥ BD.
    ∠ACB + ∠DBC = 180° [∵ Co-interior angles]
    ⇒ 90° + DBC = 180° [∵ Angle C is right angle]
    ⇒ ∠DBC = 180° – 90° = 90°
    Hence, ∠DBC is a right angle.

    (iii) In ΔDBC and ΔACB
    DB = AC [∵ ΔAMC ≅ ΔBMD]
    ∠DBC = ∠ACB [∵ Proved above]
    BC = BC [∵ M is the mid-point of line segment AB]
    Hence, ΔDBC ≅ ΔACB [∵ SAS Congruency Rule]

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