Let D, E, F be the feet of altitude from A, B, C in triangle ABC. Prove that the perpendicular bisector of ED also bisects BC.

Question

Let $D, E, F$ be the feet of altitude from $A, B, C$ in $\triangle ABC$. Prove that the perpendicular bisector of $EF$ also bisects $BC$.

Source:Challenges and Thrills of Pre College Maths

What I did:I tried by angle chasing and sine rule but I failed

I guess it is entire question. Please tell me how to improve this question or consider reopening as this is one of causes of my ban.

Answer 1

$\angle CFB=\angle BEC=90°$

Therefore $FBCE$ is a cyclic quadrilateral with diameter $BC$.

We know that the line drawn from centre of a circle to centre of a chord in it is perpendicular bisector of the chord. Let $O$ be the centre of circle($BCEF$). Perpendicular bisector from $EF$ will cross through $O$.

As proved above $BC$ is the diameter of circle($BCEF$) that means $O$ lies on it and is its midpoint. Hence the statement is true.