In the figure below, BAD,BCE,ACF and DEF are straight lines. It is given that BA=BC, AD=AF, EB=ED. If ∠BED=x∘, find the value x.
Answer:
108∘
- Given BAD,BCE,ACF and DEF are straight lines and BA=BC, AD=AF and EB=ED. Also, ∠BED=x∘.
Let ∠ABC=y∘ and ∠BAC=z∘ - In △ABC,
∠BAC=z∘=∠BCA[Given BA=BC]⟹∠BAC+∠BCA+∠ABC=180∘[By Angle sum property of triangles]⟹2z+y=180.....(1) - In △BED,
EB=ED[Given]⟹∠EBD=∠EDB=y∘[Since ∠EBD=∠ABC] - Since BAD is a straight line,
∠FAD=180−∠BAC=180−z .....(2)
In △ADF,
∠ADF=y∘=∠DFA[Given AD=AF]⟹∠ADF+∠DFA+∠FAD=180∘[By Angle sum property of triangles] ⟹2y+180−z=180 By (2) ⟹2y−z=0.....(3) - From eq(1) and (3), we get,
y=36∘
Now, in △BED,
∠BED=180−∠EBD−∠EDB⟹x=180−2y⟹x=180−2×36[Substituting the value of y]⟹x=108∘ - Hence, the value of x is 108∘.