Question:
In figure 9.38, AE bisects ∠CAD and ∠B = ∠C. Prove that AE ∥ BC.
Solution:
Let ∠B = ∠C = x
Then,
∠CAD = ∠B + ∠C = 2x (exterior angle)
⇒1/2∠CAD = x
⇒ ∠EAC = x
⇒ ∠EAC = ∠C
These are alternate interior angles for the lines AE and BC
∴ AE ∥ BC
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