Question
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.
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Answer
Given: $\triangle\text{ABC}$ 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 $\triangle\text{ABC},\ \ \text{AB}=\text{AC} [$equal sides of an isosceles triangle$]$
$\Rightarrow\angle\text{ACB}=\angle\text{ABC}\ \ ...(\text{i})$ [angles opposite to the equal sides are equal]
Since, $DE || BC$
$\Rightarrow\angle\text{ADE}=\angle\text{ACB} [$corresponding angles$] ....(ii)$
On adding both sides by $\angle\text{EDC}$ in Eq. $(ii)$, we get,
$\angle\text{ADE}+\angle\text{EDC}=\angle\text{ACB}+\angle\text{EDC}$
$\Rightarrow180^\circ=\angle\text{ACB}+\angle\text{EDC}$ [$\angle\text{ADE}$ and $\angle\text{EDC}$ from linear pair aniom]
$\Rightarrow\angle\text{EDC}+\angle\text{ABC}=180^\circ$ [from Eq. (i)]
Hence, $BCDE$ is cyclic quadrilateral, because sum opposite angles is $180^\circ .$
To prove: Quadrilateral $BCDE$ is a cyclic quadrilateral.
Construction: Draw a circle passes through the points $B, C, D$ and $E.$
Proof: In $\triangle\text{ABC},\ \ \text{AB}=\text{AC} [$equal sides of an isosceles triangle$]$
$\Rightarrow\angle\text{ACB}=\angle\text{ABC}\ \ ...(\text{i})$ [angles opposite to the equal sides are equal]
Since, $DE || BC$
$\Rightarrow\angle\text{ADE}=\angle\text{ACB} [$corresponding angles$] ....(ii)$
On adding both sides by $\angle\text{EDC}$ in Eq. $(ii)$, we get,
$\angle\text{ADE}+\angle\text{EDC}=\angle\text{ACB}+\angle\text{EDC}$
$\Rightarrow180^\circ=\angle\text{ACB}+\angle\text{EDC}$ [$\angle\text{ADE}$ and $\angle\text{EDC}$ from linear pair aniom]
$\Rightarrow\angle\text{EDC}+\angle\text{ABC}=180^\circ$ [from Eq. (i)]
Hence, $BCDE$ is cyclic quadrilateral, because sum opposite angles is $180^\circ .$
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