Question

in figure O is in the interior of triangle ABC, OD perpendicular to BC ,OE perpendicular to AB ,show that

1. OA²+OB²+OC²-OD²-OE²-OF²=AF²+BD²+CE²

2. AF²+BD²+CE²=AE²+CD²+BF²

Answer 1

HELLO DEAR,

given that:-

∆ OBC and O is a point in the enterior of the ∆ABC

OC⊥BC , OE⊥AC , and OF⊥OB

construction :- now join the point O from B,C,andA

IN ∆OAF ,<F=90°

OA²= AF²+OF²---------(1)

IN ∆OEC , <E=90°

OC²=OE²+EC²--------------(2)


in another , ∆OBD, <D=90°


OB²=OD²+BD²-----------(3)

now adding---(1) ,---(2),,and --(3)

we get,

OA²+OC²+OB²=AF²+OF²+OE²+EC²+OD²+BD²

=> OA²+OB²+OC²-OD²-OE²-OF²=AF²+BD²+CE²


(2) is in fig.




I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

hope it will help you✌✌