Question

In the given figure, Angle B = Angle D = 90° and AB = DE.
Prove that CD = BC.
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Answer 1

In ∆ABC and ∆EDC

angle BCA = angle DCE ... (vertically opposite angles)

angle ABC = angle EDC ... (each 90°)

AB = ED ... (given)

hence, by AAS congruency test,

∆ABC ≅ ∆EDC

therefore, BC = CD ... (corresponding parts of congruent triangles are congruent)

Answer 2

Answer:

given,

Angle B = Angle D = 90° and

AB = DE

Angle ACB=DC (vertically opposite angles)

thus

By ASA test (Angle Side Angle test)

triangle ABC=DCE

Thus

BC=CD

sides of an congruent triangles.

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