If a line is drawn parallel to the base


If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic.


Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC.

To prove Quadrilateral BCDE is a cyclic quadrilateral.

Construction Draw a circle passes through the points B, C, D and E.

Proof in $\triangle A B C$,  [equal sides of an isosceles triangle]

$\Rightarrow$ $\angle A C B=\angle A B C$ $\ldots(1)$

[angles opposite to the equal sides are equal]

Since, $D E \| B C$

$\Rightarrow \quad \angle A D E=\angle A C B \quad$ [corresponding angles] ...(ii)

On adding both sides by $\angle E D C$ in Eq. (ii), we get

$\angle A D E+\angle E D C=\angle A C B+\angle E D C$

$\Rightarrow \quad 180^{\circ}=\angle A C B+\angle E D C$

[ $\angle A D E$ and $\angle E D C$ form linear pair aniom]

$\Rightarrow \quad \angle E D C+\angle A B C=180^{\circ} \quad$ [from Eq. (1)]

Hence, $B C D E$ is a cyclic quadrilateral, because sum of the opposite angles is $180^{\circ}$.


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