Let- ABC be an isosceles triangle with AB - AC and let - denote its circumcircle- A point D is on arc AB of - not containing C- A point E is on arc AC of - not containing B- If AD - CE- prove that BE is parallel to AD-

We note that triangle AEC and triangle BDA are congruent. ∴ AE = nbsp;BD And hence ∠ ABE = ∠ DAB. This proves that AD is parallel to BE. ...

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Standard IX

Mathematics

Line Segment That Subtends Equal Angles at Two Other Points

Let, ABC be...

Question

Let, $A B C$ be an isosceles triangle with $A B = A C$ and let $Γ$ denote its circumcircle. A point $D$ is on arc $A B$ of $Γ$ not containing $C$ . A point $E$ is on arc $A C$ of $Γ$ not containing $B$ . If $A D = C E$ , prove that $B E$ is parallel to $A D .$

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Solution

We note that triangle AEC and triangle BDA are congruent.
$∴ A E = B D$
And hence $∠$ ABE = $∠$ DAB.
This proves that AD is parallel to BE.

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Let, ABC be an isosceles triangle with AB=AC and let Γ denote its circumcircle. A point D is on arc AB of Γ not containing C. A point E is on arc AC of Γ not containing B. If AD=CE, prove that BE is parallel to AD.

We note that triangle AEC and triangle BDA are congruent. ∴AE=BDAnd hence ∠ABE = ∠DAB. This proves that AD is parallel to BE.

Let, $A B C$ be an isosceles triangle with $A B = A C$ and let $Γ$ denote its circumcircle. A point $D$ is on arc $A B$ of $Γ$ not containing $C$ . A point $E$ is on arc $A C$ of $Γ$ not containing $B$ . If $A D = C E$ , prove that $B E$ is parallel to $A D .$

We note that triangle AEC and triangle BDA are congruent.
$∴ A E = B D$
And hence $∠$ ABE = $∠$ DAB.
This proves that AD is parallel to BE.