Question

In Fig $\left(VIII\right),ABCD$ is a rhombus . $EABF$ is a straight line such that $EA$ $=AB=BF$ , $ED$ and $FC$ when produced, meet at $O$. Prove that $\angle DOC={90}^{\circ }$

Answer 1

In $△EAD$ and $△ABC$

$\Rightarrow$ $EA=AB$ [ Given ]

$\Rightarrow$ $\angle EAD=\angle ABC$ [ Corresponding angles ]

$\Rightarrow$ $AD=BC$ [ Sides of rhombus are equal ]

$\Rightarrow$ $△EAD\cong △ABC$ [ By $SAS$ congruence rule ]

$\Rightarrow$ $\angle AED=\angle BAC$ [ By C.P.C.T ]

In $△CBF$ and $△DAB$

$\Rightarrow$ $BF=AB$ [ Given ]

$\Rightarrow$ $\angle CBF=\angle DAB$ [ Corresponding angles ]

$\Rightarrow$ $AD=BC$ [ Sides of rhombus ]

$\Rightarrow$ $△CBF\cong △DAB$ [ By $SAS$ congruence rule ]

$\Rightarrow$ $\angle CFB=\angle DBA$ [ C.P.C.T ]

In $△ABX,$

$\Rightarrow$ $\angle ABX+\angle BAX+\angle AXB={180}^o.$

$\Rightarrow$ $\angle ABX+\angle BAX={90}^o$ [ In rhombus diagonals bisect at ${90}^o.$ so, $\angle AXB={90}^o$ ]

$\Rightarrow$ $\angle CFB+\angle AED={90}^o$ [ $\angle DBA=\angle ABX=\angle CFB,\angle AED=\angle BAX=\angle BAC$ ]

$\Rightarrow$ $\angle OFE+\angle OEF={90}^o$

We have,

$\angle EOF+\angle OFE+\angle OEF={189}^o$

$\angle EOF+{90}^o={180}^o.$

$\angle EOF={90}^o.$

$\therefore$ $\angle DOC={90}^o$