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Question
In a Δ ABC, the internal bisector of angle A meets opposite side BC at point D. Through vertex C, line CE is drawn parallel to DA which meets BA produced at point E. Show that Δ ACE is isosceles.
In a Δ ABC, the internal bisector of angle A meets opposite side BC at point D. Through vertex C, line CE is drawn parallel to DA which meets BA produced at point E. Show that Δ ACE is isosceles.
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Solution
DA || CE[Given]
∠1=∠4.............1[Corresponding angles]
∠2=∠3............2
[Alternate angles]
But ∠1=∠2.............3
[ AD is the bisector of 
A]
From (i), (ii) and (iii)
∠3=∠4
AC = AE
Δ ACE is an isosceles triangle.
DA || CE[Given]
∠1=∠4.............1[Corresponding angles]
∠2=∠3............2
[Alternate angles]
But ∠1=∠2.............3
[ AD is the bisector of A]
From (i), (ii) and (iii)
∠3=∠4
AC = AE
Δ ACE is an isosceles triangle.
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