In the given figure ∆ABC is an isosceles triangle in which AB = AC and a circle passing through B and C intersects AB and AC at D and E respectively.
Question:

In the given figure ABC is an isosceles triangle in which AB = AC and a circle passing through B and C intersects AB and AC at D and E respectively. Prove that DE || BC.

Solution:

ABC is an isosceles triangle.
Here, AB = AC
∴ ∠ACB = ∠ABC   …(i)
So, exterior ∠ADE = ∠ACB 
= ∠ABC   [from(i)]
∴ ∠ADE = ∠ABC  (Corresponding angles)
Hence, DE || BC

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