JEE Advance - Mathematics (2006 - No. 27)
In $$\Delta ABC$$, internal angle bisector of $$\angle A$$ meets side $$BC$$ in $$D$$. $$DE \bot AD$$ meets $$AC$$ in $$E$$ and $$AB$$ in $$F$$. Then
$$AE$$ is $$HM$$ of $$b$$ and $$c$$
$$AD$$ $$ = {{2bc} \over {b + c}}\cos {A \over 2}$$
$$EF$$ $$ = {{4bc} \over {b + c}}\sin {A \over 2}$$
$$\Delta AEF$$ is isosceles
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