\
Pythagoras Theoram - Proving
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Question 7 Deleted for CBSE Board 2024 Exams
Question 8 Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams You are here
Last updated at April 16, 2024 by Teachoo
\
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