Question

How to do this question ?

Answer 1

i ) AM = BM   [ Since M is the mid-point ]

   DM = CM   [ Given ]

   ∠AMC = ∠BMD  [ Vertically opposite angles ]

    ∴ Δ AMC ≅ Δ BMD

ii ) Since Δ AMC ≅ Δ BMD

    ∠MAC = ∠MBD

    But they form a pair of alternate interior angles

     ∴ AC ║DB

     Now , BC is a transversal intersecting parallel lines AC and DB ,

     ∴ ∠BCA + ∠DBC = 180°

         ∠BCA = 90° [ GIVEN ]

         90° + ∠DBC = 180°

         ∠DBC = 180° - 90° = 90°

         ∠DBC = 90°

iii )  Δ AMC ≅ Δ BMD [ PROVED ]

     AC = BD [ CPCT ]

     Now , in Δ DBC and ΔACB , we have

     ∠DBC = ∠ACB  [ EACH 90° ]

      BD = CA    [ PROVED ]

       BC = CB   [ COMMON ]

     ∴ Δ DBC ≅ Δ ACB   [by SAS rule ]

iv ) Δ DBC ≅ Δ ACB

      DC = AB   [ CPCT ]

      DM = CM [ GIVEN ]

      CM = 1/2 DC = 1/2 AB

∴    CM = 1/2 AB