Question

O is a point in the interior of LABC, OD 1 BC,OE 1 AC and
OF 1 AB, as shown in the figure.
E
B
D
C
Prove that:
(1) OAP + OB2 + OCP-OD2-OE-OF- AFP - BD - CE

(ii) AF2 + BD2+CE? = AE? +BFP+CD​

Answer 1

Answer:

where is the fig. bro??

BTW here is your answer:-

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

Answer 2

$\small\bold{\underline{\sf{\blue{solution:-}}}}$

∆ 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