Question

Please answer
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Answer 1

Step-by-step explanation:

1. in triangle ABC and ADB ,

                AD = BC (given)

                AB = AB (common)

                AC = BD (diagonals of a ||gm)

       ABC ≅ ADB(sss rule)

         ∠A = ∠B (cpctc)

3rd problem is also proved.

2.||| ly , in triangle BCD and ACD ,

               AD = BC(given)

               CD = CD(common)

               AC = BD(diagonals of a ||gm)

              BCD ≅ ACD(sss rule)

             ∠C = ∠D (cpctc)

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for your understanding,

||gm - parallelogram

||| ly - similarly

Answer 2

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