Question

From a point O in the interior of a ∆ABC, perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that:
(i) AF² + BD² + CE² = OA² + OB² + OC² – OD² – OE² – OF²
(ii) 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

I HOPE ITS HELP YOU DEAR,

THANKS

Answer 2

Answer:

Explanation:

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