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In an isosceles triangle ABC with AB=AC ,a circle passing through b and c INTERSECTS THE SIDES AB AND intersects the sides AB and AC AT D and E

respectively .prove that DE is parallel to side BC



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To prove that DE is parallel to BC,

if we prove that angle ADE = Angle ABC , hence it will be proved because of corresponding angle property

so we will prove it first


∠B = ∠C   .... (1)

In the cyclic quadrilateral CBDE, side BD is produced to A.

We know that exterior angle is equal to opposite interior angle.

i.e. ∠ADE = ∠C .... (2)

From (1) and (2) –


SO corresponding angles are equal

Ans hence  DE is parrallel to BC

1 5 1
To prove: DE || BC.Proof:In order to prove that DE || BC it is sufficient to show that ∠B = ∠ADEIn ΔABC,AB = ACIn the cyclic quadrilateral CBDE, side BD is produced to A.It is the property of cyclic quadrilateral that its exterior angle is equal to the opposite interior angle.⇒ ∠ADE = ∠C .... (2)From (1) and (2), we get, which forms corresponding angles.Hence DE || BC.
2 5 2
i am sir, vikram
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