Question 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
Question8
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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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