Question

ABCD is a cyclic quadrilateral BA and CD produced, meet at E . Prove that $\Delta EBC$ and $\Delta EDA$ are equiangular.

Answer 1

$ABCD$ is a cyclic quadrilateral and $AB$ and $CD$ are produced to meet at $E.$

We know that opposite angles of a cyclic quadrilateral are supplementary.

$\therefore$ $\angle BAD+\angle BCD={180}^o$ ------ ( 1 )

and $\angle BAD+\angle EAD={180}^o$ ----- ( 2 )

From equation ( 1 ) and ( 2 ), we get,

$\angle BAD+\angle EAD=\angle BAD+\angle BCD$

$\therefore$ $\angle EAD=\angle BCD$ --- ( 3 )

Similarly,

$\Rightarrow$ $\angle ABC+\angle ADC={180}^o$ ---- ( 4 )

and $\angle ADC+\angle ADE={180}^o$ ----- ( 5 )

From equation ( 4 ) and ( 5 ), we get

$\Rightarrow$ $\angle ADC+\angle ADE=\angle ADC+\angle ABC$

$\Rightarrow$ $\angle ADE=\angle ABC$ ----- ( 6 )

And $\angle AED=\angle BEC$ [ Common angle ] ---- ( 7 )

$\therefore$ From equation ( 3 ), ( 6 ) and ( 7 ), $△EBC$ and $△EDA$ are equiangular.