. 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.
Attachments:
Answers
Answered by
4
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.
Similar questions