Question

let ABC be a triangle with AB=AC. If D is the mid-point of BC, E the foot of the perpendicular drawn from D to AC, F is the mid-point of DE. Prove that AF is perpendicular to BE.

Answer 1

Let `BC` be taken as `X`-axis with origin at `D`, the mid-point of `BC` and `DA` will be `Y`-axis.
Given, `AB=AC`
Let `BC=2alpha`, then the coordinates of `B` and `C` are `(-a,0)` and `(a,0)` let `A(0,h)`.

Then, equation of `AC` is
`{:(x,,y),(,"|",),(a,,h):}=1`
and equation of `DEbotAC` and passing through origin is
`(x)(h)-(y)/(a)=0`
`impliesx=(hy)/(a)`......`(ii)`
On solving, Eqs. `(i)` and `(ii)`, we get the coordinates of point `E` as follows
`(hy)/(a^(2))+(y)/(h)=1impliesy=(a^(2)h)/(a^(2)+h^(2))`
`:.` Coordinate of `E=((ah^(2))/(a^(2)+h^(2)),(a^(2)h)/(a^(2)+h^(2)))`
Since, `F` is mid-point of `DE`.
`:.` Coordinate of `F[(ah^(2))/(2(a^(2)+h^(2))),(a^(2)h)/(2(a^(2)+h^(2)))]`
`:.` Slope of `AF`,
`m_(1)=(h-(a^(2)h)/(2(a^(2)+h^(2))))/(0-(ah^(2))/(2(a^(2)+h^(2))))=(2h(a^(2)+h^(2))-a^(2)h)/(-ah^(2))`
`impliesm_(1)=(-(a^(2)+h^(2)))/(ah)` .....`(iii)`
and slope of `BE` , `m_(2)=((a^(2)h)/((a^(2)+h^(2)))-0)/((ah^(2))/((a^(2)+h^(2)))+a)=(a^(2)h)/(ah^(2)+a^(3)+ah^(2))`
`impliesm_(2)=(ah)/(a^(2)+2h^(2))`........`(iv)`
From Eqs. `(iii)` and `(iv)`, `m_(1)m_(2)=-1impliesAFbotBE`