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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 - Mathematics

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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 + CE= OA2 + OB2 + OC2 - OD2 - OE2 - OF2

Sum

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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Chapter 17: Pythagoras Theorem - Exercise 17.1

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Frank Mathematics [English] Class 9 ICSE
Chapter 17 Pythagoras Theorem
Exercise 17.1 | Q 14.1

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