Question

$ABCD$ is rhombus. If $AB$ is produced at $F$ and $BA$ is produced at $E$ such that $AB=AE=BF$, then result will be ?

• A. $ED||CF$
• B. $ED>CF$
• C. $ED\perp CF$
• D. ${ED}^2+{CF}^2={EF}^2$

Answer 1

The correct option is C $ED\perp CF$

Produce $EC$ and $FC$ to meet at a point $M.$

Let each side of the rhombus be $x.$ Then,

$\Rightarrow$ $AB=BC=CD=AD=BF=AE=x$

$\Rightarrow$ $\angle DAB+\angle CBA={180}^o$

$\Rightarrow$ $\angle DAE+\angle CBF={180}^o$

In $△ADF,$

$\Rightarrow$ $\angle EAD+\angle AED+\angle ADE={180}^o$ ------ ( 1 )

In $△BCF,$

$\Rightarrow$ $\angle CBF+\angle CFB+\angle BCF={180}^o$ ----- ( 2 )

$\Rightarrow$ $\angle AED=\angle ADE$ --- ( 3 ) [ Base angles of an equal sides are also equal ]

$\Rightarrow$ $\angle CFB=\angle BCF$ ---- ( 4 ) [ Base angles of an equal sides are also equal ]

Adding equation ( 1 ) and ( 2 ),

$\Rightarrow$ $(\angle EAD+\angle CBF)+(\angle AED+\angle ADE)+(\angle CFB+\angle BCF)={360}^o$

$\Rightarrow$ ${180}^o+2\angle AED+2\angle CFB={360}^o$ [ From ( 3 ) and ( 4 ) ]

$\Rightarrow$ $\angle AED+\angle CFB={90}^o.$

$\Rightarrow$ $\angle EMF={90}^o$

$\Rightarrow$ $EM\perp FM$

$\therefore$ $ED\perp CF$