In an isosceles triangle ABC with AB = AC a circle passing through B and C intersect the sides AB and AC at D and E respectively.Prove that DE||BC.
To prove: DE || BC.
In order to prove that DE || BC it is sufficient to show that ∠B = ∠ADE
AB = AC
In 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.