From a point O inside a triangle ABC, perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB, respectively. Prove that the perpendiculars from A, B and C to the sides EF, FD and DE are concurrent

Explanation

Let the position vectors of points A,B,C,D,E and F be `veca,vecb, vecc, vecd ,vece and vecf` w.r.t O. Let perpendiculars from A to EF and from B to DF meet each other at H. let position vectors of H be `vecr`. We join CH. In order to prove the statement given in the question. it is sufficient it is sufficient to prove that CH is prependicular to DE.
Now `as OD bot BC Rightarrow vecd.(vecb-vecc)=0`
`Rightarrow vecd.vecb = vecd.vecc`
`as OE botAC Rightarrowvece.(vecc-veca)=0 Rightarrow vece.vecc=vece.veca`
as `OF bot ABRightarrow vecf. (veca-vecb)=0 Rightarrow vecf.veca=vecf.vecb`
Also `AH bot EF Rightarrow (vecr.veca) . (vece-vecf)=0`
` Rightarrow vecr.vece -vecr.vecf-veca.vece + veca.vecf=0`
`and BH bot FD Rightarrow (vecr-vecb) . (vecf-vecd)=0`
`Rightarrow vecr.vecf-vecr.vecd-vecb.vecf+vecb.vecd=0`
Adding (iv) and (v) , we get
`vecr.vece-veca.vece+veca.vecf-vecr.vecd-vecb.vecf+vecb.vecd=0`
`or vecr.(vece.vecd)- vece.vecc+vecd.vecc=0` [ using (i), (ii) and (iii))]
` or (vecr - vecc) . (vece -vecd) =0`
`Rightarrow vec(CH). vec(ED) = 0 Rightarrow CH bot ED `