In the figure below, ^@BAD, BCE, ACF^@ and ^@DEF^@ are straight lines. It is given that ^@BA = BC^@, ^@AD = AF^@, ^@EB = ED^@. If ^@\angle BED = x^\circ^@, find the value ^@x^@.
A B C D E F


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Answer:

^@108^\circ^@

Step by Step Explanation:
  1. Given ^@BAD, BCE, ACF^@ and ^@DEF^@ are straight lines and ^@BA = BC^@, ^@AD = AF^@ and ^@EB = ED^@. Also, ^@\angle BED = x^\circ.^@
    Let ^@\angle ABC = y^\circ^@ and ^@\angle BAC = z^\circ^@
  2. In ^@\triangle ABC^@,
    ^@\begin{align} & \angle BAC = z^\circ = \angle BCA && [\text{Given } BA = BC] \\ \implies & \angle BAC + \angle BCA + \angle ABC = 180^\circ && [\text{By Angle sum property of triangles}]\\ \implies & 2z + y = 180 && .....(1) \end{align}^@
  3. In ^@\triangle BED^@,
    ^@\begin{align} & EB = ED && \text{[Given]}\\ \implies & \angle EBD = \angle EDB = y^\circ && [\text{Since } \angle EBD = \angle ABC] \end{align}^@
  4. Since ^@BAD^@ is a straight line,
    ^@\angle FAD = 180 - \angle BAC = 180 - z \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space .....(2)^@
    In ^@\triangle ADF^@,
    ^@\begin{align} & \angle ADF = y^\circ = \angle DFA && [\text{Given } AD = AF]\\ \implies & \angle ADF + \angle DFA + \angle FAD = 180^\circ && [\text{By Angle sum property of triangles}] \space\space\space\\ \implies & 2y + 180 - z = 180 && \text{ By (2) }\\ \implies & 2y - z = 0 && .....(3) \end{align}^@
  5. From eq^@(1)^@ and ^@(3)^@, we get,
    ^@y = 36^\circ^@
    Now, in ^@\triangle BED^@,
    ^@\begin{align} & \angle BED = 180 - \angle EBD - \angle EDB \\ \implies & x = 180 - 2y \\ \implies & x = 180 - 2 \times 36 && [\text{Substituting the value of } y] \\ \implies & x = 108^\circ \end{align}^@
  6. Hence, the value of ^@x^@ is ^@108^\circ^@.

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