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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 = OA2 + OB2 + OC2 - OD2 - OE2 - OF2
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Solution
In right triangle OFA, ODB and OEC, we have
OA2 = AF2 + OF2
OB2 = BD2 + OD2
OC2 = CE2 + OE2
Adding all these results, we get
OA2 + OB2 + OC2 = AF2 + BD2 + CE2 + OF2 + OD2 + OE2
⇒ AF2 + BD2 + CE2 = OA2 + OB2 + OC2 - OD2 - OE2 - OF2.
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