Ex 6.5, 8
In figure, O is a point in the interior of a triangle ABC, OD ⊥BC, OE ⊥ AC and OF ⊥AB. Show that
OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
Given: Triangle ABC and
O is a point in the interior of a
triangle ABC where,
OD ⊥𝐵𝐶,𝑂𝐸⊥𝐴𝐶,𝑂𝐹⊥𝐴𝐵
To prove :- OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
Proof:-
Let us join the point O from A , B and C.
Using Pythagoras theorem,
(Hypotenuse)2 = (Height)2 + (Base)2
In a right angle triangle OAF.
(OA)2 = AF2 + OF2
In right angle triangle ODB
OB2 = OD2 + BD2
In a right angle triangle OEC
(OC)2 = (OE)2 + (EC)2
Adding (1) + (2) + (3)
(OA)2 + (OB)2 + (OC)2 = AF2 + OF2 + OD2 + BD2 + OE2 + EC2
OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
Hence proved
Ex 6.5,8
In figure, O is a point in the interior of a triangle ABC, OD ⊥BC, OE ⊥AC and OF ⊥AB. Show that
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
Using Pythagoras theorem.
(Hypotenuse)2 = (Height)2 + (Base)2
In Δ ODB, OB2 = OD2 + BD2
In Δ OFB, OB2 = OF2 + FB2
In Δ OFA, OA2 = OF2 + AF2
In Δ OEA, OA2 = OE2 + AE2
In Δ OEC, OC2 = OE2 + CE2
In Δ ODC, OC2 = OD2 + CD2

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.