ABC is an isoscles triangle in which AB=AC. A circle passing through B and C intersects the sides AB and AC at D and E respectively then show that DE is parallel to BC?

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ABC is an isoscles triangle in which AB=AC. A circle passing through B and C intersects the sides AB and AC at D and E respectively then show that DE is parallel to BC?

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Answer 1 · 2018-06-01T01:52:21

see explanation.

Explanation:

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Given $A B = A C, \Rightarrow ∠ A B C = ∠ A C B$ $AB=AC, => angleABC=angleACB$
Let $∠ A B C = x, \Rightarrow ∠ A C B = x$ $angleABC=x, => angleACB=x$
As $B C E D$ $BCED$ is a cyclic quadrilateral, opposite angles add up to $180^{\circ}$ $180^@$ ,
$\Rightarrow ∠ B D E = 180 - ∠ E C B = 180 - x$ $=> angleBDE=180-angleECB=180-x$ ,
$\Rightarrow A D E = 180 - ∠ B D E = 180 - (180 - x) = x$ $=> ADE=180-angleBDE=180-(180-x)=x$
as $∠ A D E = ∠ A B C = x$ $angleADE=angleABC=x$ ,
$\Rightarrow D E$ $=> DE$ // $B C$ $BC$

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ABC is an isoscles triangle in which AB=AC. A circle passing through B and C intersects the sides AB and AC at D and E respectively then show that DE is parallel to BC?

1 Answer

Explanation:

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