. In an acute angled triangle ABC, the internal bisector of angle A meets base BC at point D. DE I AB and DF I AC; show that AEF is an isosceles triangle.
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ABC is an acute angled triangle, the internal bisector of angle A meets base BC at point D.
- DE ⊥ AB
- DF ⊥ AC
- AEF is an isosceles triangle.
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