Question

In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that (i) OA 2 + OB 2 + OC 2 − OD 2 − OE 2 − OF 2 = AF 2 + BD 2 + CE 2 (ii) AF 2 + BD 2 + CE 2 = AE 2 + CD 2 + BF 2

Answer 1

Answer:

$In\:\triangle\:OAF,\:OA^2=OF^2+AF^2$ ............(1)

$In\:\triangle\:OAD,\:OB^2=OD^2+BD^2$ .............(2)

$In\:\triangle\:OCE,\:OC^2=OE^2+CE^2$ .............(3)

Adding (1),(2) and (3)

$OA^2+OB^2+OC^2=OF^2+AF^2+OD^2+BD^2+OE^2+CE^2$

$\implies\boxed{OA^2+OB^2+OC^2-OD^2-OE^2-OF^2=AF^2+BD^2+CE^2}$

...........................(4)

$In\:\triangle\:OEA,\:OA^2=OE^2+AE^2$..................(5)

$In\:\triangle\:OFB,\:OB^2=OF^2+BF^2$..................(6)

$In\:\triangle\:ODC,\:OC^2=OD^2+CD^2$..................(7)

Using (5),(6) and (7 ) in (4)

$(OE^2+AE^2)+(OF^2+BF^2)+(OD^2+CD^2)-OD^2-OE^2-OF^2+AF^2+BD^2+CE^2$

$AE^2+BF^2+CD^2=AF^2+BD^2+CE^2$

$\implies\boxed{AF^2+BD^2+CE^2=AE^2+BF^2+CD^2}$