Question

In the figure, ABCD is a square in which CE and DE are the angle bisector of <C and < D, prove that <CED = <A=<B=<C=
<D.​

Answer 1

Answer:

∠CED = ∠A = ∠B = ∠C = ∠D = 90°

Step-by-step explanation:

E is the center point of the Square. The line drawn from the D and C to E is parallel to each other. And also, if the line drawn from the corner point (D and C) to the center point (E) of the square, the angle produced by them is 45°. Because ∠D = ∠C = 90°. This 90° exactly bisects by a line, which made 45°.

So, ∠CDE = ∠DCE = 45°

The Sum of three interior angles of triangle is 180°

∠CDE + ∠DCE + ∠CED = 180°

45° + 45° + ∠CED = 180°

∠CED = 90°

The ABCD is a square. So, all corner angles should be 90°

∠A = ∠B = ∠C = ∠D = 90°      ∵[∠A+∠B+∠C+∠D = 360°]

So, ∠CED = ∠A = ∠B = ∠C = ∠D = 90°