Question

In triangle ABC, AB = AC. A circle passing through B and c intersects the sides AB and AC at D and E respectively. Prove that DE || BC. 

Answer 1

To prove = DE II BC 

Proof: In cydic quadrilateral DECB 

∠ DEC + ∠ DBC = 80°  - ( 1) (Opposite angles of cyclic quadrilateral) 

Also, ∠ AED + ∠ DEC = 80° - (2) (Linear pair) 

From (1) and (2), we get, 

∠ DBC = ∠ AED - (3) 

AB= AC (given) 

∴ ∠ABC = ∠ ACB - (4) (angles opposite to equal sides of triangle) 

From (3) and ( 4) ⇒  ∠ AED = ∠ ACB 

But, these are corresponding angles 

∴ DE II BC 

Answer 2

In order to prove that DE || BC, it is sufficient to show that ∠B = ∠ADE.
In Δ ABC, we have
AB = AC ⇒ ∠B = ∠C       ...(i)

In the cyclic quadrilateral CBDE, side BD is produced to A.
∴ ∠ADE = ∠C        ...(ii)( ∵ Exterior angle = Opposite interior angle )

From (i) and (ii), we get
∠B = ∠ADE.
Hence, DE || BC.             ...Hence proved.