Question

$ABCD$ is a parallelogram. The circle through $A,B$ and $C$ intersects $CD$ produced at $E$. If $AB=10cm,BC=8cm,CE=14cm$ . Find $AE$

Answer 1

Given : $ABCD$ is a parallelogram and a circle is drawn through $A,B$ and $C$ which intersects $CD$ produced, at $E$

$AB=10cm$ $BC=8cm$ $CE=14cm$

Now,

$\angle AED+\angle ABC={180}^0\to \left(1\right)$ (Sum of opposite angles of a cyclic quadrilateral is supplementary)

$\angle EDA+\angle ADC={180}^0\to \left(2\right)$ (linear pair)

$\angle EDA+\angle ABC={180}^0\to \left(3\right)$ ($\because$ $\angle ADC=\angle ABC$ as they are the opposite angles of a parallelogram)

From $\left(1\right)$ & $\left(3\right)$ we have,

$\angle AED=\angle EDA\to \left(4\right)$

$\Rightarrow$ in $\Delta AED$ two angles are equal, hence the opposite sides corresponding to these angles must also be equal

i.e. $AE=AD$

and $AD=BC$ (opposite sides of a parallelogram are equal)

$\therefore$ $AE=8cm$.