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
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
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