In the figure below, ^@BAD, BCE, ACF^@ and ^@DEF^@ are straight | Math Question and Answer

Answer:

$108^\circ$

Step by Step Explanation:

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 $\angle BAC = z^\circ$
In $\triangle ABC$ ,
$\begin{align} & \angle BAC = z^\circ = \angle BCA && [\text{Given } BA = BC] \\ \implies & \angle BAC + \angle BCA + \angle ABC = 180^\circ && [\text{By Angle sum property of triangles}]\\ \implies & 2z + y = 180 && .....(1) \end{align}$
In $\triangle BED$ ,
$\begin{align} & EB = ED && \text{[Given]}\\ \implies & \angle EBD = \angle EDB = y^\circ && [\text{Since } \angle EBD = \angle ABC] \end{align}$
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 $\triangle ADF$ ,
$\begin{align} & \angle ADF = y^\circ = \angle DFA && [\text{Given } AD = AF]\\ \implies & \angle ADF + \angle DFA + \angle FAD = 180^\circ && [\text{By Angle sum property of triangles}] \space\space\space\\ \implies & 2y + 180 - z = 180 && \text{ By (2) }\\ \implies & 2y - z = 0 && .....(3) \end{align}$
From eq $(1)$ and $(3)$ , we get,
$y = 36^\circ$
Now, in $\triangle BED$ ,
$\begin{align} & \angle BED = 180 - \angle EBD - \angle EDB \\ \implies & x = 180 - 2y \\ \implies & x = 180 - 2 \times 36 && [\text{Substituting the value of } y] \\ \implies & x = 108^\circ \end{align}$
Hence, the value of $x$ is $108^\circ$ .

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