Answer:
[tex]\angle BED = 135^\circ[/tex]
Step-by-step explanation:
First of all, let us do a construction in the given question image.
Let us connect the points E and C.
Please refer to the image attached in the answer area.
Step 1:
EC is drawn in the diagram.
Now, consider the triangles △EGC and △EGB.
1. Side BG = GC (G is the point on perpendicular bisector of AC)
2. [tex]\angle EGB =\angle EGC[/tex] (G is the point on perpendicular bisector of AC)
3. Line EG is common to both the triangles.
So,from SAS (Side Angle Side) congruence i.e. two sides and angle between them are equal in two triangles.
△EGC ≅ △EGB
Step 2:
Now, we know that △EGC ≅ △EGB, we can say that the corresponding sides CE and BE must be equal to each other i.e. CE [tex]\cong[/tex] BE .
And we are already given that BC [tex]\cong[/tex] BE.
[tex]\therefore[/tex] In [tex]\triangle BEC[/tex]:
EB ≅ EC ≅ BC that means [tex]\triangle BEC[/tex] is an equilateral triangle.
Hence, every angle of [tex]\triangle BEC[/tex] must be equal to [tex]60^\circ[/tex].
[tex]\therefore[/tex] ∠BEC = 60°.
Step 3:
Let us find ∠ECD= ?
ABCD is a square, so [tex]\angle BCD =90^\circ[/tex]
[tex]\angle BCE + \angle ECD = 90^\circ\\\Rightarrow 60^\circ+ \angle ECD = 90^\circ\\\Rightarrow \angle ECD = 30^\circ[/tex]
Step 4:
We know that ABCD is a square, so side BC = CD.
We have proved in the previous step that BC = EC
[tex]\therefore[/tex] in [tex]\triangle CED[/tex], two sides are equal so it is isosceles triangle.
One angle ∠ECD= [tex]30^\circ[/tex]
As it is an isosceles triangle, the other two angles will be equal.
Let them be equal to [tex]x^\circ[/tex].
The sum of all 3 angles of a triangle is equal to [tex]180^\circ[/tex]
[tex]x+x+30=180\\\Rightarrow x = 75^\circ[/tex]
OR
we can say [tex]\angle CED =75^\circ[/tex]
Now, from the diagram, we can see the following:
[tex]\angle BED = \angle BEC + \angle CED[/tex]
[tex]\Rightarrow \angle BED = 60^\circ+75^\circ\\\Rightarrow \angle BED = 135^\circ[/tex]