Question 8:

In the following 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

Answer:

Join OA, OB, and OC.

(i) Applying Pythagoras theorem in ∆AOF, we obtain

OA² = OF² + AF²

Similarly, in ∆BOD,

OB² = OD² + BD²

Similarly, in ∆COE,

OC² = OE² + EC²

Adding these equations,

OA² + OB² + OC² = OF² + AF² + OD² + BD² + OE² + EC²

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

(ii) From the above result,

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

∴ AF² + BD² + EC² = AE² + CD² + BF²

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