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Question
From a point O in the interior of aΔABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that: AF2 + BD2 + CE2 = AE2 + CD2 + BF2
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Solution
In right triangles ODB and ODC, we have
OB2 = OD2 + BD2
OC2 = OD2 + CD2
∴ OB2 - OC2 = (OD2 + BD2) - (OD2 + CD2)
⇒ OB2 - OC2 = BD2 - CD2 ....(i)
Similarly, we have
OC2 - OA2 = CE2 - AE2 ....(ii)
OA2 - OB2 = AF2 - BF2 ....(iii)
Adding (i), (ii) and (iii), we get
(OB2 - OC2) + (OC2 - OA2) + (OA2 - OB2) = (BD2 - CD2) + (CE2 - AE2) + (AF2 - BF2)
⇒ (BD2 + CE2 + AF2) - (AE2 + CD2 + BF2) = 0
⇒ AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
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