Question

in an isosceles triangle ABC with AB equal to AC a circle passing through B and C intersect side AC and ab at D and e respectively prove that d e parallel BC​

Answer 1

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

In ΔABC,

∠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) –

∠ADE = ∠ABC

So corresponding angles are equal

Ans: Hence  DE is parrallel to BC.

Answer 2

$\LARGE\mathfrak{\underline{\underline{\red{To~prove: }}}}$

DE || BC.Proof:In order to prove that DE || BC it is sufficient to show that ∠B = ∠ADE ln 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.