Question

In figure O is a point in the interior of the triangle ABC, OD is perpendicular to BC, OE is perpendicular to AC and OF perpendicular to AB. Show that $A{F}^2+B{D}^2+C{E}^2=A{E}^2+C{D}^2+B{F}^2$

Answer 1

Join $OB,OA$ and $OC$

In right angled $△OAF,$

$\Rightarrow$ $O{A}^2=A{F}^2+O{F}^2$ ---- ( 1 ) [ By Pythagoras theorem ]

In right angled $△OBD,$

$\Rightarrow$ $O{B}^2=O{D}^2+B{D}^2$ ---- ( 2 ) [ By Pythagoras theorem ]

In right angled $△OCE,$

$\Rightarrow$ $O{C}^2=O{E}^2+C{E}^2$ ---- ( 3 ) [ By Pythagoras theorem ]

In right angled $△OAE,$

$\Rightarrow$ $O{A}^2=O{E}^2+A{E}^2$ ----- ( 4 ) [ By Pythagoras theorem ]

In right angled $△OCD,$

$\Rightarrow$ $O{C}^2=O{D}^2+C{D}^2$ ------ ( 5 ) [ By Pythagoras theorem ]

In right angled $△OFB,$

$\Rightarrow$ $O{B}^2=O{F}^2+B{F}^2$ ------ ( 6 ) [ By Pythagoras theorem ]

$A{F}^2+B{D}^2+C{E}^2=(O{A}^2-O{F}^2)+(O{B}^2-O{D}^2)+(O{C}^2-O{E}^2)$ [ From ( 1 ), ( 2 ) and ( 3 ) ]

$=O{A}^2+O{B}^2+O{C}^2-O{F}^2-O{D}^2-O{E}^2$

$A{E}^2+C{D}^2+B{F}^2=(O{A}^2-O{E}^2)+(O{C}^2-O{D}^2)+(O{B}^2-O{F}^2)$ [ From ( 4 ), ( 5 ) and ( 6 ) ]

$=O{A}^2+O{B}^2+O{C}^2-O{F}^2-O{D}^2-O{E}^2$

Therefore, we have proved that,

$Rightarrow$ $A{F}^2+B{D}^2+C{E}^2=A{E}^2+C{D}^2+B{F}^2$