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[Solved] In Figure, O is a point in the interior of a triangle. ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.

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In Figure, O is a point in the interior of a triangle.

ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that:

(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2

(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.

1 Answer
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Given, in ΔABC, O is a point in the interior of a triangle.

And OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.

Join OA, OB and OC

(i) By Pythagoras theorem in ΔAOF, we have

OA2 = OF2 + AF2

Similarly, in ΔBOD

OB2 = OD2 + BD2

Similarly, in ΔCOE

OC2 = OE2 + EC2

Adding these equations,

OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE+ EC2

OA2 + OB2 + OC2 – OD2 – OE2 – OF2 

= AF2 + BD2 + CE2.

(ii) AF2 + BD2 + EC2 = (OA2 – OE2) + (OC2 – OD2) + (OB2 – OF2)

∴ AF2 + BD2 + CE2 

= AE2 + CD2 + BF2.

This post was modified 3 years ago by Raavi Tiwari
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