Question

in the figure BC = BD and ray DE ||ray CB,Ray BE bisects angle A B D .prove :

i)angle EBD= BDC
ii)⬜BCDE is a parallelogram​

Answer 1

Since, these angles that is angles EBD and BDC are also alternate angles, we can say that, CD is parallel to BE. Also, BCDE is a parallelogram.

Here, we first need to prove that the triangles CBD and BDE are congruent to each other.

Now, in triangles CBD and BDE, we have that,

BD = BD ...{∵ Common part of both the triangles}

BC = BD ...{∵ Given in the problem}

∠EDB = ∠DBC  ...{alternate angles}

Hence, by SAS axiom of congruency, we get that,

Triangles CBD and BDE are congruent to each other.

Then, we get, consequently, angles EBD and BDC are equal since they are corresponding parts of congruent triangles.

Since, these angles that is angles EBD and BDC are also alternate angles, we can say that, CD is parallel to BE.

Hence, BCDE is a parallelogram.

Hence, proved.

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